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Why does relativistic velocity addition sometimes produce a larger velocity than classical velocity addition?
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I got a question for homework saying that there were two rockets on a parallel track, heading towards earth. Rocket A was in front of Rocket B. Rocket A was travelling at a velocity of $0.75c$ from the frame of reference (FOR) of earth and Rocket B was travelling at a velocity of $0.5c$ from the FOR of earth. The question asked to find the velocity of Rocket A from the FOR of Rocket B.
I got an answer of $0.4c$,
$$v_AB=fracv_a-v_b1-v_av_b/c^2=frac0.75c-0.5c1-0.75ccdot0.5c/c^2=0.4c,$$
which both my teachers said was right.
I am confused because this velocity is greater than what I got using the formula for classical velocity addition ($0.25c$). For every other question I have done, the relativistic velocity is less than the classical velocity. I am wondering why the relativistic velocity is larger in this case?
homework-and-exercises special-relativity reference-frames velocity
edited 59 mins ago
Kyle Kanos
21.4k114792
asked 2 hours ago
Lachie G
83
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Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
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I got a question for homework saying that there were two rockets on a parallel track, heading towards earth. Rocket A was in front of Rocket B. Rocket A was travelling at a velocity of $0.75c$ from the frame of reference (FOR) of earth and Rocket B was travelling at a velocity of $0.5c$ from the FOR of earth. The question asked to find the velocity of Rocket A from the FOR of Rocket B.
I got an answer of $0.4c$,
$$v_AB=fracv_a-v_b1-v_av_b/c^2=frac0.75c-0.5c1-0.75ccdot0.5c/c^2=0.4c,$$
which both my teachers said was right.
I am confused because this velocity is greater than what I got using the formula for classical velocity addition ($0.25c$). For every other question I have done, the relativistic velocity is less than the classical velocity. I am wondering why the relativistic velocity is larger in this case?
homework-and-exercises special-relativity reference-frames velocity
edited 59 mins ago
Kyle Kanos
21.4k114792
asked 2 hours ago
Lachie G
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I got a question for homework saying that there were two rockets on a parallel track, heading towards earth. Rocket A was in front of Rocket B. Rocket A was travelling at a velocity of $0.75c$ from the frame of reference (FOR) of earth and Rocket B was travelling at a velocity of $0.5c$ from the FOR of earth. The question asked to find the velocity of Rocket A from the FOR of Rocket B.
I got an answer of $0.4c$,
$$v_AB=fracv_a-v_b1-v_av_b/c^2=frac0.75c-0.5c1-0.75ccdot0.5c/c^2=0.4c,$$
which both my teachers said was right.
I am confused because this velocity is greater than what I got using the formula for classical velocity addition ($0.25c$). For every other question I have done, the relativistic velocity is less than the classical velocity. I am wondering why the relativistic velocity is larger in this case?
homework-and-exercises special-relativity reference-frames velocity
edited 59 mins ago
Kyle Kanos
21.4k114792
asked 2 hours ago
Lachie G
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I got a question for homework saying that there were two rockets on a parallel track, heading towards earth. Rocket A was in front of Rocket B. Rocket A was travelling at a velocity of $0.75c$ from the frame of reference (FOR) of earth and Rocket B was travelling at a velocity of $0.5c$ from the FOR of earth. The question asked to find the velocity of Rocket A from the FOR of Rocket B.
I got an answer of $0.4c$,
$$v_AB=fracv_a-v_b1-v_av_b/c^2=frac0.75c-0.5c1-0.75ccdot0.5c/c^2=0.4c,$$
which both my teachers said was right.
I am confused because this velocity is greater than what I got using the formula for classical velocity addition ($0.25c$). For every other question I have done, the relativistic velocity is less than the classical velocity. I am wondering why the relativistic velocity is larger in this case?
homework-and-exercises special-relativity reference-frames velocity
homework-and-exercises special-relativity reference-frames velocity
edited 59 mins ago
Kyle Kanos
21.4k114792
asked 2 hours ago
Lachie G
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 59 mins ago
Kyle Kanos
21.4k114792
asked 2 hours ago
Lachie G
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 59 mins ago
Kyle Kanos
21.4k114792
edited 59 mins ago
Kyle Kanos
21.4k114792
edited 59 mins ago
Kyle Kanos
21.4k114792
21.4k114792
asked 2 hours ago
Lachie G
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Lachie G
83
asked 2 hours ago
Lachie G
83
83
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lachie G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.
Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.
answered 1 hour ago
Wolphram jonny
10.2k22451
add a comment |Â
up vote
1
down vote
My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:
“Alice sees Bob move at velocity $u~hat x$ and Bob sees Carol move at velocity $v~hat x$, how fast does Alice see Carol move?â€Â
The answer here is to construct the world line in Bob’s reference frame, $textCarol=(ct, vt)_textBob, text for all t$, then boost it by velocity $-uhat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_textAlice= c~fracgamma_u~(vt+beta_uct)gamma_u~(ct+beta_uvt)=fracv+u1+uv/c^2.$$
But now you are confronted instead with the problem,
“Alice sees Bob move at velocity $u~hat x$ and Carol move at velocity $v~hat x$, how fast does Bob see Carol move?â€Â
Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-uhat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+uhat x$ hence the sign on $u$ has changed to give you,
$$v_textBob= c~fracgamma_u~(vt-beta_uct)gamma_u~(ct-beta_uvt)=fracv-u1-uv/c^2.$$
Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.
answered 2 hours ago
CR Drost
20.8k11758
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.
Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.
answered 1 hour ago
Wolphram jonny
10.2k22451
add a comment |Â
up vote
3
down vote
accepted
If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.
Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.
answered 1 hour ago
Wolphram jonny
10.2k22451
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.
Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.
answered 1 hour ago
Wolphram jonny
10.2k22451
If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.
Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.
answered 1 hour ago
Wolphram jonny
10.2k22451
edited 1 hour ago
answered 1 hour ago
Wolphram jonny
10.2k22451
answered 1 hour ago
Wolphram jonny
10.2k22451
answered 1 hour ago
Wolphram jonny
10.2k22451
10.2k22451
add a comment |Â
add a comment |Â
up vote
1
down vote
My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:
“Alice sees Bob move at velocity $u~hat x$ and Bob sees Carol move at velocity $v~hat x$, how fast does Alice see Carol move?â€Â
The answer here is to construct the world line in Bob’s reference frame, $textCarol=(ct, vt)_textBob, text for all t$, then boost it by velocity $-uhat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_textAlice= c~fracgamma_u~(vt+beta_uct)gamma_u~(ct+beta_uvt)=fracv+u1+uv/c^2.$$
But now you are confronted instead with the problem,
“Alice sees Bob move at velocity $u~hat x$ and Carol move at velocity $v~hat x$, how fast does Bob see Carol move?â€Â
Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-uhat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+uhat x$ hence the sign on $u$ has changed to give you,
$$v_textBob= c~fracgamma_u~(vt-beta_uct)gamma_u~(ct-beta_uvt)=fracv-u1-uv/c^2.$$
Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.
answered 2 hours ago
CR Drost
20.8k11758
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
add a comment |Â
up vote
1
down vote
My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:
“Alice sees Bob move at velocity $u~hat x$ and Bob sees Carol move at velocity $v~hat x$, how fast does Alice see Carol move?â€Â
The answer here is to construct the world line in Bob’s reference frame, $textCarol=(ct, vt)_textBob, text for all t$, then boost it by velocity $-uhat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_textAlice= c~fracgamma_u~(vt+beta_uct)gamma_u~(ct+beta_uvt)=fracv+u1+uv/c^2.$$
But now you are confronted instead with the problem,
“Alice sees Bob move at velocity $u~hat x$ and Carol move at velocity $v~hat x$, how fast does Bob see Carol move?â€Â
Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-uhat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+uhat x$ hence the sign on $u$ has changed to give you,
$$v_textBob= c~fracgamma_u~(vt-beta_uct)gamma_u~(ct-beta_uvt)=fracv-u1-uv/c^2.$$
Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.
answered 2 hours ago
CR Drost
20.8k11758
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:
“Alice sees Bob move at velocity $u~hat x$ and Bob sees Carol move at velocity $v~hat x$, how fast does Alice see Carol move?â€Â
The answer here is to construct the world line in Bob’s reference frame, $textCarol=(ct, vt)_textBob, text for all t$, then boost it by velocity $-uhat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_textAlice= c~fracgamma_u~(vt+beta_uct)gamma_u~(ct+beta_uvt)=fracv+u1+uv/c^2.$$
But now you are confronted instead with the problem,
“Alice sees Bob move at velocity $u~hat x$ and Carol move at velocity $v~hat x$, how fast does Bob see Carol move?â€Â
Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-uhat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+uhat x$ hence the sign on $u$ has changed to give you,
$$v_textBob= c~fracgamma_u~(vt-beta_uct)gamma_u~(ct-beta_uvt)=fracv-u1-uv/c^2.$$
Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.
answered 2 hours ago
CR Drost
20.8k11758
My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:
“Alice sees Bob move at velocity $u~hat x$ and Bob sees Carol move at velocity $v~hat x$, how fast does Alice see Carol move?â€Â
The answer here is to construct the world line in Bob’s reference frame, $textCarol=(ct, vt)_textBob, text for all t$, then boost it by velocity $-uhat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_textAlice= c~fracgamma_u~(vt+beta_uct)gamma_u~(ct+beta_uvt)=fracv+u1+uv/c^2.$$
But now you are confronted instead with the problem,
“Alice sees Bob move at velocity $u~hat x$ and Carol move at velocity $v~hat x$, how fast does Bob see Carol move?â€Â
Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-uhat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+uhat x$ hence the sign on $u$ has changed to give you,
$$v_textBob= c~fracgamma_u~(vt-beta_uct)gamma_u~(ct-beta_uvt)=fracv-u1-uv/c^2.$$
Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.
answered 2 hours ago
CR Drost
20.8k11758
answered 2 hours ago
CR Drost
20.8k11758
answered 2 hours ago
CR Drost
20.8k11758
answered 2 hours ago
CR Drost
20.8k11758
20.8k11758
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
add a comment |Â
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
What does the x with the ^ symbol mean? Also, thankyou for the answer.
– Lachie G
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
That's a unit vector in the $x$-direction. A 3D vector like displacement or velocity can be specified in terms of three components $v_x,y,z$, denoting the combination variously as $vec v = v_x~hat x +v_y~hat y +v_z~hat z,$ or you sometimes see $mathbf v = v_x~mathbf i + v_y~mathbf j + v_z~mathbf k,$ or even just $$mathbf v=v_xbeginbmatrix1\0\0endbmatrix+v_ybeginbmatrix0\1\0endbmatrix+v_zbeginbmatrix0\0\1endbmatrix=beginbmatrixv_x\v_y\v_zendbmatrix.$$ Vector algebra is the proper way to articulate that a quantity has both a magnitude and a direction.
– CR Drost
1 hour ago
add a comment |Â
Lachie G is a new contributor. Be nice, and check out our Code of Conduct.
Lachie G is a new contributor. Be nice, and check out our Code of Conduct.
Lachie G is a new contributor. Be nice, and check out our Code of Conduct.
Lachie G is a new contributor. Be nice, and check out our Code of Conduct.
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