Solving an equation with vector coefficients

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I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$

Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
3 hours ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
3 hours ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
3 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$

Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
3 hours ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
3 hours ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
3 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$

Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$

Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?

equation-solving tensors algebra

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

edited 10 mins ago

m_goldberg

82.5k869190

edited 10 mins ago

m_goldberg

82.5k869190

edited 10 mins ago

m_goldberg

82.5k869190

asked 3 hours ago

Sean McAllister

1134

New contributor

asked 3 hours ago

Sean McAllister

1134

asked 3 hours ago

Sean McAllister

1134

New contributor

Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
3 hours ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
3 hours ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
3 hours ago

add a commentÂ |Â

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
3 hours ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
3 hours ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
3 hours ago

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
3 hours ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
3 hours ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
3 hours ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

addendum

If you want preserve the invariant scalarproducts, the equation could be defined as follows

Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

add a commentÂ |Â

up vote
3
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 3 hours ago

Carl Woll

60k279154

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

addendum

If you want preserve the invariant scalarproducts, the equation could be defined as follows

Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

add a commentÂ |Â

up vote
4
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

addendum

If you want preserve the invariant scalarproducts, the equation could be defined as follows

Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

add a commentÂ |Â

up vote
4
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

addendum

If you want preserve the invariant scalarproducts, the equation could be defined as follows

Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

addendum

If you want preserve the invariant scalarproducts, the equation could be defined as follows

Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

edited 2 hours ago

answered 3 hours ago

Ulrich Neumann

5,080413

answered 3 hours ago

Ulrich Neumann

5,080413

answered 3 hours ago

Ulrich Neumann

5,080413

add a commentÂ |Â

up vote
3
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 3 hours ago

Carl Woll

60k279154

add a commentÂ |Â

up vote
3
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 3 hours ago

Carl Woll

60k279154

add a commentÂ |Â

up vote
3
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 3 hours ago

Carl Woll

60k279154

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 3 hours ago

Carl Woll

60k279154

answered 3 hours ago

Carl Woll

60k279154

answered 3 hours ago

Carl Woll

60k279154

answered 3 hours ago

Carl Woll

60k279154

add a commentÂ |Â

Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.

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