# [Special Relativity] Test Particle inside Sun's Gravitational Field

## Homework Statement:

[B]Question[/B]

[Question Context: Consider the motion of a test particle of (constant) mass $m$ inside the gravitational field produced by the sun in the context of special relativity.]

Consider the equations of motion for the test particle, which can be written as $$\frac{d(m\gamma c)}{dt} = \frac{\vec{v}}{c} \cdot \vec{F},$$

OR

$$\frac{d(m\gamma \vec{v})}{dt} = \vec{F},$$

where $\vec{v}$ is the speed of the test particle, $c$ is the (constant) speed of light, and by definition, $$\gamma \equiv \frac{1}{\sqrt{1- \frac{\vec{v}^2}{c^2}}} .$$

In addition, the gravitational force is given by $$\vec{F} \equiv -\frac{GMm}{r^2} \hat{e}_r$$

where $\hat{e}_r$ is the unit vector in the direction between the Sun (of mass M) and the test particle (of mass $m$).

Now, integrate the first equation above - that is, $\frac{d(m\gamma c)}{dt} = \frac{\vec{v}}{c} \cdot \vec{F}$ - to find $\gamma$ as a function of $r$, by using the property that $$-\frac{\dot{r}}{r^2} = \frac{d}{dt} \Big(\frac{1}{r}\Big).$$

You may need to introduce a constant of integration. This will be a free parameter of the solution.

## Homework Equations:

Please refer below $\Longrightarrow$
Below is an attempted solution based off of another user's work on StackExchange:
Source: [https://physics.stackexchange.com/questions/525169/special-relativity-test-particle-inside-the-suns-gravitational-field/525212#525212]

To begin with, I will be using the following equation mentioned in the question - that is, $$\vec{F} \equiv -\frac{GMm}{r^2} \hat{e}_r.$$

Using this, I get the following by plugging the equation for gravitational force to the other one shown above in the question section:
$$\frac{d(m\gamma c)}{dt}=-\frac{\dot{r}}{c}\cdot \frac{GMm}{r^2} \hat{e}_r$$

where both $m$ and $c$ are constants. With this in mind, I could pull those two constants out of the deferential before rewriting the given formula to:

$$c^2\frac{d\gamma}{dt}=GMm\frac{d}{dt}\left(\frac{1}{r}\right) \hat{e}_r$$

Now, I can integrate both sides to get the following answer:

$$c^2\gamma=\frac{GMm}{r} \hat{e}_r+k,$$
where $k$ the constant of integration.

Comment
As mentioned before, this is a solution that is largely based upon the work by another used named "fielder". However, after looking through the problem, did this user succeeded in finding $\gamma$ as a function of $r$? If the user is correct in his approach, however, why is that?

Therefore, to put it simply, if the attempted solution is incorrect, any amount of guidance to help me reach the correct answer is much appreciated. However, if the attempted solution is correct, I would greatly appreciate it if anybody here on the forum could briefly explain why the attempted solution above succeeded in finding $\gamma$ as a function of $r$.

Thank you for reading through this and I would sincerely appreciate any amount of assistance!

Last edited:

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Orodruin
Staff Emeritus
Homework Helper
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Are you trying to use Newtonian gravity applied to SR? That does not work very well.

Below is an attempted solution based off of another user's work on StackExchange:
As mentioned before, this is a solution that is largely based upon the work by another used named "fielder".
Do not refer to other material without providing a proper reference: In this case a link to the SE thread.

Are you trying to use Newtonian gravity applied to SR? That does not work very well.

Do not refer to other material without providing a proper reference: In this case a link to the SE thread.
Thank you for your reply. In that case, if not Newtonian gravity, what would be a better alternative to solve for the question?

On the side note, I have included the SE source. Thank you for the reminder and I apologize for forgetting to put it here in the first place. Beyond that, the stated question on SE is more or less the same here.

Ibix
Thank you for your reply. In that case, if not Newtonian gravity, what would be a better alternative to solve for the question?
Special relativity (which limits causal effects to light speed) is incompatible with Newtonian gravity (where gravity propagates at infinite speed - something not even unambiguously defined in relativity). You need GR. Look up Sean Carroll's lecture notes online - the correct equations of motion are around equation 7.38, from memory.

PeroK
Homework Helper
Gold Member
$$c^2\gamma=\frac{GMm}{r} \hat{e}_r+k,$$
Apart from anything else, this equation is dimensionally and vectorially wrong.

Special relativity (which limits causal effects to light speed) is incompatible with Newtonian gravity (where gravity propagates at infinite speed - something not even unambiguously defined in relativity). You need GR. Look up Sean Carroll's lecture notes online - the correct equations of motion are around equation 7.38, from memory.
Thank you for the helpful comment. I sincerely appreciate it!
However, after looking up Sean Carroll's lecture notes online, I fail to recognize a GR equation of motion near equation 7.38. Beyond that, though, the class I am taking is an SR class and we have learned little to nothing about GR. Perhaps it's just me, but it seems odd that such a question needing to involve GR is handed out to students in an SR class.

However, if this is - oddly - the only way to solve the problem, would it be possible if you could help me point out specifically what equation are you exactly referring to?

PeroK
Homework Helper
Gold Member
However, after looking up Sean Carroll's lecture notes online, I fail to recognize a GR equation of motion near equation 7.38. Beyond that, though, the class I am taking is an SR class and we have learned little to nothing about GR. Perhaps it's just me, but it seems odd that such a question needing to involve GR is handed out to students in an SR class.

However, if this is - oddly - the only way to solve the problem, would it be possible if you could help me point out specifically what equation are you exactly referring to?

You'll need to ask whoever is taking the class about this. Mathematically, of course, there is nothing to stop you taking the equations you have been given and trying to solve them.

Apart from anything else, this equation is dimensionally and vectorially wrong.
Thank you for the clarification! After a bit more studying around and asking questions, the above calculation process (as well as the "answer") is indeed incorrect.

Below is part of the question-answering process provided by another user in SE (same link). And, from what I am able to tell, the equations do indeed look familiar. However, despite the user's "explanation", I fail to understand he means exactly.

Here's the potential solution:

First, force is always the rate of momentum change:
$$F = \frac{d(m_0 \gamma_v \vec{v})}{dt} = m_0 \frac{d\gamma_v}{dt} \vec{v} + m_0 \gamma_v \vec{a}$$

and the rate of change of the kinetic energy is
$$\frac{dE_k}{dt} = \vec{v} \cdot \vec{F}$$

noting that

$$\vec{v} = \frac{d\vec{r}}{dt}$$

In addition to that, note that $E = \gamma_v m_0 c^2 \Longrightarrow \frac{d\gamma_v}{dt} = \frac{\vec{v} \cdot \vec{F}}{m_0 c^2}$

-------------------------

However, after seeing the "solution" above, I fail to see how this would help me answer the provided question - that is, "to find $\gamma$ as a function of $r$".

If possible, I would sincerely appreciate any assistance or clarification you can provide. Once again, thank you for pointing the error behind the first provided "solution".

PeroK
Homework Helper
Gold Member
Thank you for the clarification! After a bit more studying around and asking questions, the above calculation process (as well as the "answer") is indeed incorrect.

Below is part of the question-answering process provided by another user in SE (same link). And, from what I am able to tell, the equations do indeed look familiar. However, despite the user's "explanation", I fail to understand he means exactly.

Here's the potential solution:

First, force is always the rate of momentum change:
$$F = \frac{d(m_0 \gamma_v \vec{v})}{dt} = m_0 \frac{d\gamma_v}{dt} \vec{v} + m_0 \gamma_v \vec{a}$$

and the rate of change of the kinetic energy is
$$\frac{dE_k}{dt} = \vec{v} \cdot \vec{F}$$

noting that

$$\vec{v} = \frac{d\vec{r}}{dt}$$

In addition to that, note that $E = \gamma_v m_0 c^2 \Longrightarrow \frac{d\gamma_v}{dt} = \frac{\vec{v} \cdot \vec{F}}{m_0 c^2}$

-------------------------

However, after seeing the "solution" above, I fail to see how this would help me answer the provided question - that is, "to find $\gamma$ as a function of $r$".

If possible, I would sincerely appreciate any assistance or clarification you can provide. Once again, thank you for pointing the error behind the first provided "solution".
Is this supposed to be for a particle in free fall, initially at rest wrt the Sun?

Is this supposed to be for a particle in free fall, initially at rest wrt the Sun?
The question did not specifically state that. According to the question, we should "consider the motion of a test particle of (constant) mass $m$ inside the gravitational field produced by the Sun in the context of special relativity". And, nothing else from that. However, from my point of view, I see nothing implying a free-falling particle initially at rest wrt the Sun.

Do you perhaps have any thoughts regarding this?

PeroK
Homework Helper
Gold Member
The question did not specifically state that. According to the question, we should "consider the motion of a test particle of (constant) mass $m$ inside the gravitational field produced by the Sun in the context of special relativity". And, nothing else from that. However, from my point of view, I see nothing implying a free-falling particle initially at rest wrt the Sun.

Do you perhaps have any thoughts regarding this?
You could at least try the simple case first. If we assume a simple radial plunge scenario, then it's not too hard to get $$F = \gamma^3 ma$$ where $a$ is the coordinate acceleration. Perhaps try to prove this first. Note that this is generally true for 1D motion in SR.

Hint: try differentiating $\gamma$. And, be very careful! It's easy to go wrong.

It we take it at face value, this gives us:$$\gamma^3 \ddot r = - \frac{GM}{r^2}$$
Where $\gamma^3 = (1- \frac{\dot r^2}{c^2})^{-3/2}$.

Can you solve this differential equation?

To begin with, I went ahead and tried going about solving (or, more accurately, proving) $F = \gamma^3 ma$. To see the calculation process as well as the source I turned to for help, here is a site I used for my reference when doing the calculation: https://www.physicsforums.com/threads/show-that-f-gamma-3-ma.338744/.

It we take it at face value, this gives us:
γ3¨r=−GMr2γ3r¨=−GMr2​
\gamma^3 \ddot r = - \frac{GM}{r^2} Where γ3=(1−˙r2c2)−3/2γ3=(1−r˙2c2)−3/2\gamma^3 = (1- \frac{\dot r^2}{c^2})^{-3/2}.

Can you solve this differential equation?
I tried to solve for the differential equation, however, I got to admit that this is beyond me - even though I should probably know this before accepting my professor's recommendation of taking the SR course. I apologize for my lack of mathematical competency. However, if you do not mind, could you please go through the steps in accomplishing this provided that you have the time and are willing to do so.

Other than that, I should probably begin looking for courses teaching differential equations specifically next semester ...

Once again, thank you for the insightful response and help earlier!

PeroK
Homework Helper
Gold Member
There's a basic technique which is to look for exact derivatives. In this case we have:
$$(1- \frac{\dot r^2}{c^2})^{-3/2} \ddot r = - \frac{GM}{r^2}$$
You were given a hint in the problem that $\frac{d}{dt}(\frac 1 r) = -\frac{\dot r}{r^2}$. So let's multiply that equation by $\dot r$.
Next, I asked you to differentiate $\gamma$. If you do this, you can relate $\frac{d\gamma}{dt}$ to $\gamma^3$ and the equation above simplifies to an exact derivative on both sides.

If you find that too hard, then this problem is probably beyond you.

Ibix
However, if this is - oddly - the only way to solve the problem, would it be possible if you could help me point out specifically what equation are you exactly referring to?
Sorry - 7.38 is part of the derivation. 7.43, 7.44, 7.47 and 7.48 are the relevant equations of motion. As PeroK says, you can of course do the maths you are being asked to do. It's not valid physics, though - if it were that easy we wouldn't bother with GR! Maybe it's an acceptable approximation in some particular regime?

Orodruin
Staff Emeritus