# Integration of an acceleration formula involving vectors

## Homework Statement

Suppose a constant force F acts on a particle of mass m initially at rest.

(a) Integrate the formula for acceleration $$\vec{a} = \frac{\vec F}{\gamma m} - \frac{\vec v}{\gamma mc^2}(\vec F \cdot \vec v)$$ where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ to show that the speed of the particle after time t is given by $$\frac{v}{c} = \frac{(F/m)t}{\sqrt{(F/m)^2t^2+c^2}}$$

Given above

## The Attempt at a Solution

I have taken Calc. 1-3, differential equations, and linear algebra, and yet I still do not know how to integrate the formula for the acceleration $\vec a$ shown above. I don't think that I've ever integrated vectors, and especially not dot products; so I have no idea where to even start. This is part of a special relativity section for an bachelor's level Astrophysics course. Any help would be greatly appreciated!
Thanks!

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haruspex
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Since F is a constant vector, it can be factored out of the integral: $\int \vec F.\vec v = \vec F. \int \vec v$.

Since F is a constant vector, it can be factored out of the integral: $\int \vec F.\vec v = \vec F. \int \vec v$.
Yes, but now I guess I would have (not sure about the dotted $\vec v$ 's in the integral):
$$\int \vec a \, dt = \vec v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt$$ and I still don't know how to evaluate this with vectors (especially the dot product in the integral), and the equation I'm trying to get to (the v/c equation) has just scalars in it, so I don't know how to get there from all these vectors even if I knew how to evaluate that integral.

haruspex
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$$\frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt$$
No, as you suspected, that's not valid.
Thing to notice is that you are asked to find speed, not velocity, yet you start with acceleration as a vector.
This suggests that the integration to be performed starts with a scalar on the left, so you have to do something before integrating. Any ideas?

No, as you suspected, that's not valid.
Thing to notice is that you are asked to find speed, not velocity, yet you start with acceleration as a vector.
This suggests that the integration to be performed starts with a scalar on the left, so you have to do something before integrating. Any ideas?
Well, the only thing that comes to mind would be breaking the vectors into component form, but I'm not exactly sure how to do that with this specific setup. I tried (as an attempt) just making all the vectors scalars and then integrating, but then I got an equation involving v and r (position), and the v/c equation does not involve position. Also, I'm not sure that was the correct thing to do anyway.

haruspex
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Well, the only thing that comes to mind would be breaking the vectors into component form, but I'm not exactly sure how to do that with this specific setup. I tried (as an attempt) just making all the vectors scalars and then integrating, but then I got an equation involving v and r (position), and the v/c equation does not involve position. Also, I'm not sure that was the correct thing to do anyway.
No, that's not what I had in mind. What's simple equation relating v, as a speed, to the corresponding velocity? What do you get if you differentiate?

No, that's not what I had in mind. What's simple equation relating v, as a speed, to the corresponding velocity? What do you get if you differentiate?
I'm not sure I understand what you mean, here's what I think you mean by the simple equation relating $v$ to $\vec v$ (correct me if I'm wrong):
$$v = |\vec v|$$

haruspex
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I'm not sure I understand what you mean, here's what I think you mean by the simple equation relating $v$ to $\vec v$ (correct me if I'm wrong):
$$v = |\vec v|$$
Almost. Slightly simpler (removes the modulus signs) if you square both sides: $v^2 = \vec v.\vec v$. Now differentiate.

Almost. Slightly simpler (removes the modulus signs) if you square both sides: $v^2 = \vec v.\vec v$. Now differentiate.
Ok, so do you mean to replace $$v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt$$ with this $$v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int v^2 \, dt$$? If so, how do I make the force vector just the force?

Ok, so do you mean to replace $$v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int \vec v \cdot \vec v \, dt$$ with this $$v = \frac{\vec F}{\gamma m}t - \frac{\vec F}{\gamma mc^2} \int v^2 \, dt$$? If so, how do I make the force vector just the force?

Almost. Slightly simpler (removes the modulus signs) if you square both sides: $v^2 = \vec v.\vec v$. Now differentiate.
$$\frac{d}{dt}[\vec v \cdot \vec v] = \frac{d}{dt}[v^2] = 2v\frac{dv}{dt}$$

haruspex
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$$\frac{d}{dt}[\vec v \cdot \vec v] = \frac{d}{dt}[v^2] = 2v\frac{dv}{dt}$$
Right, but you can also perform the derivative on the left, to get an expression involving vectors, velocity and acceleration.

Right, but you can also perform the derivative on the left, to get an expression involving vectors, velocity and acceleration.
On the left of which equation? (Also, thanks for helping me out so far)

haruspex
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On the left of which equation? (Also, thanks for helping me out so far)
Perform the derivative $\frac d{dt} (\vec v . \vec v)$, without converting to scalars first. Use the product rule.

Perform the derivative $\frac d{dt} (\vec v . \vec v)$, without converting to scalars first. Use the product rule.
Alright, I get
$$\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v$$

haruspex
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Alright, I get
$$\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v$$
Right, so how can you use that with your starting equation?

Right, so how can you use that with your starting equation?
I'm sorry, but I am not sure. Are you talking about the problem's equation for $\vec a$? If so, everything I try doesn't really get me anywhere.

haruspex
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I'm sorry, but I am not sure. Are you talking about the problem's equation for $\vec a$? If so, everything I try doesn't really get me anywhere.
Yes. What do you have to do to turn the left hand side into $\frac d{dt} (\vec v . \vec v)$?

Yes. What do you have to do to turn the left hand side into $\frac d{dt} (\vec v . \vec v)$?
The only thing that comes to my mind is solving $\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v$ for $\vec a$ and substituting it in the starting equation, but this involves "dividing" by a vector, which doesn't make sense. I am sorry that I can't figure this out very easily (I am easily confused).

haruspex
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The only thing that comes to my mind is solving $\frac{d}{dt}[\vec v \cdot \vec v] = 2\vec a \cdot \vec v$ for $\vec a$
No, you need to do something to both sides of the equation so that the left hand side becomes $\vec a \cdot \vec v$. Nothing complicated.

Btw, it seems to me you have to make use of the information that not only is the force constant but the particle starts from rest. What does that tell you about the relationship between the directions of the vectors, F, v, a?

No, you need to do something to both sides of the equation so that the left hand side becomes $\vec a \cdot \vec v$. Nothing complicated.

Btw, it seems to me you have to make use of the information that not only is the force constant but the particle starts from rest. What does that tell you about the relationship between the directions of the vectors, F, v, a?
Then maybe I need to dot both sides of the equation with velocity and multiply both sides by 2?

Unless my logic is wrong, F, v, and a should all be in the same direction if it starts from rest.

haruspex
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Then maybe I need to dot both sides of the equation with velocity and multiply both sides by 2?

Unless my logic is wrong, F, v, and a should all be in the same direction if it starts from rest.
Yes and yes.
I found it useful to work from both ends of the problem. As well taking the step mentioned above, work backward by differentiating the final equation (after first squaring both sides).

Yes and yes.
I found it useful to work from both ends of the problem. As well taking the step mentioned above, work backward by differentiating the final equation (after first squaring both sides).
Ok, so here is how far I got using that info:
$$\vec a = \frac{\vec F}{\gamma m} - \frac{\vec F}{\gamma mc^2}(\vec F \cdot \vec v)$$
$$2\vec a \cdot \vec v = \frac{2\vec F \cdot \vec v}{\gamma m} - \frac{2\vec v \cdot \vec v}{\gamma mc^2}(\vec F \cdot \vec v)$$
$$\frac{d}{dt}[\vec v \cdot \vec v] = \frac{2\vec F \cdot \vec v}{\gamma m} (1-\frac{v^2}{c})$$
$$v^2 = \int \frac{2\vec F \cdot \vec v}{\gamma m} (1-\frac{v^2}{c})\, dt$$
Since F and v are parallel, $\vec F \cdot \vec v = Fv\cos (0) = Fv$
$$v^2 = \frac{2F}{\gamma m} \int (v-\frac{v^3}{c})\, dt$$
But, I am stuck here because the first term in the integral is going to give me position, which is not in the v/c formula, and I don't think I know how to integrate $v^3$ with respect to t. What should I do? Also, the v/c formula involves t, but so far I don't have any t's (except for the integral).

haruspex
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$\frac{d}{dt}[\vec v \cdot \vec v] = \frac{2\vec F \cdot \vec v}{\gamma m} (1-\frac{v^2}{c})$
c2
$v^2 = \frac{2F}{\gamma m} \int (v-\frac{v^3}{c})\, dt$
Can't take the gamma outside the integral - it's not constant.
When you get stuck going forwards, do as I did and start working back from the end.
For me, the two met in the middle with $\vec v \cdot \vec a = \frac{Fv}{m\gamma^3}$.

c2
Can't take the gamma outside the integral - it's not constant.
When you get stuck going forwards, do as I did and start working back from the end.
For me, the two met in the middle with $\vec v \cdot \vec a = \frac{Fv}{m\gamma^3}$.
Crap, didn't think about the gamma, and totally forgot to keep the square on that c; this is what happens when I've been up since 7:15 AM (1:33 AM now), went to university, and have been doing hw problems several at a time for the rest of the day, lol.

From the end, do you mean from the v/c equation? If so, I will keep your expression in mind and try working backwards tomorrow, and if I'm still stuck (or if I figure it out), I'll post a reply. Thank you for the amount you've helped me today, and hopefully I'll reply with something tomorrow.