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Expressing the conversion from units of time to units of space in special relativity rigorously
Clash Royale CLAN TAG#URR8PPP
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In their book Spacetime Physics, Taylor and Wheeler stress the analogy of the speed of light $c$ to a conversion factor $k$ which converts between miles and meters and subsequently convert freely between units of length and units of time . In analogy to something like
$$
Delta x = 2 ,mi = 2 ,mi cdot 1600 ,fracmmi = 3800,m
$$
we get something like
$$
Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracms = 3 cdot 10^10 ,m tag1
$$
Calculating like this is very convenient and powerful (especially if the calculations involve spacetime intervals) but the symbols can't have their usual meaning because ordinary seconds can't be measured in ordinary meters.
My questions are:
- How does the meaning of the symbols $Delta t$, $m$, $s$ in equation $(1)$ differ from their ordinary meaning?
- How would I rewrite equation $(1)$ into a rigorous expression which uses the symbols in their ordinary meaning?
Or is there something wrong with equation $(1)$? As written above, it works well in calculations.
I have read the answers to some related questions (1, 2). They suggest that $Delta t$ may be a redefinition of time as the length $c Delta t$. This makes sense but it doesn't explain why units of both length and time occur in equation $(1)$.
special-relativity spacetime units
asked 53 mins ago
Marc
567312
add a comment |Â
up vote
1
down vote
favorite
In their book Spacetime Physics, Taylor and Wheeler stress the analogy of the speed of light $c$ to a conversion factor $k$ which converts between miles and meters and subsequently convert freely between units of length and units of time . In analogy to something like
$$
Delta x = 2 ,mi = 2 ,mi cdot 1600 ,fracmmi = 3800,m
$$
we get something like
$$
Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracms = 3 cdot 10^10 ,m tag1
$$
Calculating like this is very convenient and powerful (especially if the calculations involve spacetime intervals) but the symbols can't have their usual meaning because ordinary seconds can't be measured in ordinary meters.
My questions are:
- How does the meaning of the symbols $Delta t$, $m$, $s$ in equation $(1)$ differ from their ordinary meaning?
- How would I rewrite equation $(1)$ into a rigorous expression which uses the symbols in their ordinary meaning?
Or is there something wrong with equation $(1)$? As written above, it works well in calculations.
I have read the answers to some related questions (1, 2). They suggest that $Delta t$ may be a redefinition of time as the length $c Delta t$. This makes sense but it doesn't explain why units of both length and time occur in equation $(1)$.
special-relativity spacetime units
asked 53 mins ago
Marc
567312
Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In their book Spacetime Physics, Taylor and Wheeler stress the analogy of the speed of light $c$ to a conversion factor $k$ which converts between miles and meters and subsequently convert freely between units of length and units of time . In analogy to something like
$$
Delta x = 2 ,mi = 2 ,mi cdot 1600 ,fracmmi = 3800,m
$$
we get something like
$$
Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracms = 3 cdot 10^10 ,m tag1
$$
Calculating like this is very convenient and powerful (especially if the calculations involve spacetime intervals) but the symbols can't have their usual meaning because ordinary seconds can't be measured in ordinary meters.
My questions are:
- How does the meaning of the symbols $Delta t$, $m$, $s$ in equation $(1)$ differ from their ordinary meaning?
- How would I rewrite equation $(1)$ into a rigorous expression which uses the symbols in their ordinary meaning?
Or is there something wrong with equation $(1)$? As written above, it works well in calculations.
I have read the answers to some related questions (1, 2). They suggest that $Delta t$ may be a redefinition of time as the length $c Delta t$. This makes sense but it doesn't explain why units of both length and time occur in equation $(1)$.
special-relativity spacetime units
asked 53 mins ago
Marc
567312
In their book Spacetime Physics, Taylor and Wheeler stress the analogy of the speed of light $c$ to a conversion factor $k$ which converts between miles and meters and subsequently convert freely between units of length and units of time . In analogy to something like
$$
Delta x = 2 ,mi = 2 ,mi cdot 1600 ,fracmmi = 3800,m
$$
we get something like
$$
Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracms = 3 cdot 10^10 ,m tag1
$$
Calculating like this is very convenient and powerful (especially if the calculations involve spacetime intervals) but the symbols can't have their usual meaning because ordinary seconds can't be measured in ordinary meters.
My questions are:
- How does the meaning of the symbols $Delta t$, $m$, $s$ in equation $(1)$ differ from their ordinary meaning?
- How would I rewrite equation $(1)$ into a rigorous expression which uses the symbols in their ordinary meaning?
Or is there something wrong with equation $(1)$? As written above, it works well in calculations.
I have read the answers to some related questions (1, 2). They suggest that $Delta t$ may be a redefinition of time as the length $c Delta t$. This makes sense but it doesn't explain why units of both length and time occur in equation $(1)$.
special-relativity spacetime units
special-relativity spacetime units
asked 53 mins ago
Marc
567312
asked 53 mins ago
Marc
567312
edited 33 mins ago
asked 53 mins ago
Marc
567312
asked 53 mins ago
Marc
567312
asked 53 mins ago
Marc
567312
567312
Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago
add a comment |Â
Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago
Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago
add a comment |Â
2 Answers
2
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oldest
votes
up vote
3
down vote
A more rigorous form of the equation $(1)$ would be
$$ Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracmls = 3 cdot 10^10,m cdotdfracsls$$
Where $ls$ is light-second. Then for convenience one can drop $dfracsls$, but it remains implied for the dimensional analysis.
answered 31 mins ago
safesphere
6,73611239
add a comment |Â
up vote
1
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There is no difference in the symbols $Delta t$, $m$, and $s$ from their ordinary meaning.
What is different is the assertion that $3:10^8 m/s=1$ or equivalently that $1s=3:10^8 m$. The second still retains its original SI definition as does the meter. All that is changed is that a second is a certain number of meters. This means that you can measure distances in seconds or times in meters, where seconds and meters retain their SI definitions.
What does change from ordinary meaning is that length and duration have the same dimensionality. The dimensionality of a quantity is a matter of convention, so that convention can be altered as desired, but the ordinary meaning is to consider them to have different dimensionality.
answered 28 mins ago
Dale
1,857415
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
A more rigorous form of the equation $(1)$ would be
$$ Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracmls = 3 cdot 10^10,m cdotdfracsls$$
Where $ls$ is light-second. Then for convenience one can drop $dfracsls$, but it remains implied for the dimensional analysis.
answered 31 mins ago
safesphere
6,73611239
add a comment |Â
up vote
3
down vote
A more rigorous form of the equation $(1)$ would be
$$ Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracmls = 3 cdot 10^10,m cdotdfracsls$$
Where $ls$ is light-second. Then for convenience one can drop $dfracsls$, but it remains implied for the dimensional analysis.
answered 31 mins ago
safesphere
6,73611239
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A more rigorous form of the equation $(1)$ would be
$$ Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracmls = 3 cdot 10^10,m cdotdfracsls$$
Where $ls$ is light-second. Then for convenience one can drop $dfracsls$, but it remains implied for the dimensional analysis.
answered 31 mins ago
safesphere
6,73611239
A more rigorous form of the equation $(1)$ would be
$$ Delta t = 100 ,s = 100 ,s cdot 3 cdot 10^8 ,fracmls = 3 cdot 10^10,m cdotdfracsls$$
Where $ls$ is light-second. Then for convenience one can drop $dfracsls$, but it remains implied for the dimensional analysis.
answered 31 mins ago
safesphere
6,73611239
answered 31 mins ago
safesphere
6,73611239
answered 31 mins ago
safesphere
6,73611239
answered 31 mins ago
safesphere
6,73611239
6,73611239
add a comment |Â
add a comment |Â
up vote
1
down vote
There is no difference in the symbols $Delta t$, $m$, and $s$ from their ordinary meaning.
What is different is the assertion that $3:10^8 m/s=1$ or equivalently that $1s=3:10^8 m$. The second still retains its original SI definition as does the meter. All that is changed is that a second is a certain number of meters. This means that you can measure distances in seconds or times in meters, where seconds and meters retain their SI definitions.
What does change from ordinary meaning is that length and duration have the same dimensionality. The dimensionality of a quantity is a matter of convention, so that convention can be altered as desired, but the ordinary meaning is to consider them to have different dimensionality.
answered 28 mins ago
Dale
1,857415
add a comment |Â
up vote
1
down vote
There is no difference in the symbols $Delta t$, $m$, and $s$ from their ordinary meaning.
What is different is the assertion that $3:10^8 m/s=1$ or equivalently that $1s=3:10^8 m$. The second still retains its original SI definition as does the meter. All that is changed is that a second is a certain number of meters. This means that you can measure distances in seconds or times in meters, where seconds and meters retain their SI definitions.
What does change from ordinary meaning is that length and duration have the same dimensionality. The dimensionality of a quantity is a matter of convention, so that convention can be altered as desired, but the ordinary meaning is to consider them to have different dimensionality.
answered 28 mins ago
Dale
1,857415
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is no difference in the symbols $Delta t$, $m$, and $s$ from their ordinary meaning.
What is different is the assertion that $3:10^8 m/s=1$ or equivalently that $1s=3:10^8 m$. The second still retains its original SI definition as does the meter. All that is changed is that a second is a certain number of meters. This means that you can measure distances in seconds or times in meters, where seconds and meters retain their SI definitions.
What does change from ordinary meaning is that length and duration have the same dimensionality. The dimensionality of a quantity is a matter of convention, so that convention can be altered as desired, but the ordinary meaning is to consider them to have different dimensionality.
answered 28 mins ago
Dale
1,857415
There is no difference in the symbols $Delta t$, $m$, and $s$ from their ordinary meaning.
What is different is the assertion that $3:10^8 m/s=1$ or equivalently that $1s=3:10^8 m$. The second still retains its original SI definition as does the meter. All that is changed is that a second is a certain number of meters. This means that you can measure distances in seconds or times in meters, where seconds and meters retain their SI definitions.
What does change from ordinary meaning is that length and duration have the same dimensionality. The dimensionality of a quantity is a matter of convention, so that convention can be altered as desired, but the ordinary meaning is to consider them to have different dimensionality.
answered 28 mins ago
Dale
1,857415
answered 28 mins ago
Dale
1,857415
answered 28 mins ago
Dale
1,857415
answered 28 mins ago
Dale
1,857415
1,857415
add a comment |Â
add a comment |Â
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Why can't you just view equation 1 as the distance light travels in that time interval?
– Aaron Stevens
43 mins ago
Yes, that's an intuitive meaning of the equation. But how would you express this rigorously using an equation with well-defined symbols?
– Marc
34 mins ago