[Special Relativity] Test Particle Inside the Sun's Gravitational Field

[Athenian]

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.

Relevant Equations

Please refer below $\Longrightarrow$

Below is an attempted solution based off of another user's work on StackExchange:
Source: [https://physics.stackexchange.com/q...de-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!

 

[Orodruin]

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.

 

[Athenian]

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]

$$c^2\gamma=\frac{GMm}{r} \hat{e}_r+k,$$

Apart from anything else, this equation is dimensionally and vectorially wrong.

 

[Athenian]

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?

Once again, thank you for your kind assistance and helpful comment.

 

[PeroK]

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?

Once again, thank you for your kind assistance and helpful comment.

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.

 

[Athenian]

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]

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?

 

[Athenian]

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]

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?

 

[Athenian]

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]

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]

If this is a problem seriously posed in a course on SR you should ask for a refund of whatever course fee you have paid.

 

[Athenian]

@Orodruin
At this point in time, it would not be surprising for people helping me on the forum to know that I am clearly having much trouble from my SR course - hence why I can be at times incredibly active on this forum. According to my SR professor per my inquiry at the beginning of the semester, this SR course is suitable for students who just finished their first semester in university physics (learning Newtonian mechanics) and have a mild understanding of calculus (at the time I began taking the course, I just finished learning Calculus II, i.e. before even knowing partial derivatives or gradients). And, after trying my best studying the course for the semester, I have come to realize - contrary to the professor's assumptions - that clearly the course is too advanced for me (specifically, starting from the third week into the SR course).

Of course, I am by no means stating that my SR professor is a bad professor. Rather, he is an incredibly smart and great professor - but, perhaps he's more suitable for a more advanced and educated audience in my opinion.

If this is a problem seriously posed in a course on SR you should ask for a refund of whatever course fee you have paid.

To put it shortly, I honestly wouldn't mind withdrawing from the course if I had that option readily available. However, considering that I am studying abroad as an exchange student for a year, policies state that withdrawing from a course at this point is not allowed. And, even if it were and I did so, I would consequently be forced to graduate a year later at my home university and, in turn, lose my full scholarship as a result. Needless to say, if I continue with the course and fail it, I also lose my full scholarship as well.

Therefore, the dilemma is as follows:
I - somehow - forcefully withdraw the course and lose my scholarship or continue with the course and lose my scholarship as well provided that I fail it.

Under this circumstance, I decided that still trying my best with this course and attempt to pass it would be my only best option to potentially keep my scholarships and graduate on time.

Therefore, since I have joined the Physics Forums, I really appreciate the science advisors and experts alike who have been providing me their assistance with my problems on SR. Even though I may seem really clueless at times, I nevertheless really appreciate everybody's help. Thank you!

 

[Athenian]

@Ibix

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?

Since the course began, I have oftentimes been using "less conventional" solution methods to solve my assignment questions. And, at the end of the day, the professor only simply cares that one gets the answer correctly - regardless whether the method used was taught in the lecture or not.

Therefore, with that said, I have taken a look at the equations you provided. And, I have to say, I am completely lost. When it comes to dealing with GR, it would be no exaggeration to say that I would need a step-to-step explanation on how it is done in relation to the question - provided that it is allowed on the forum.

Regardless, I will try my best and see what I can do with the provided equations. Of course, if you could provide additional assistance in helping me understand how I could use the following equations to "find $\gamma$ as a function of $r$", I would sincerely appreciate it.

Thank you very much!

 

[PeroK]

@Orodruin
At this point in time, it would not be surprising for people helping me on the forum to know that I am clearly having much trouble from my SR course - hence why I can be at times incredibly active on this forum. According to my SR professor per my inquiry at the beginning of the semester, this SR course is suitable for students who just finished their first semester in university physics (learning Newtonian mechanics) and have a mild understanding of calculus (at the time I began taking the course, I just finished learning Calculus II, i.e. before even knowing partial derivatives or gradients). And, after trying my best studying the course for the semester, I have come to realize - contrary to the professor's assumptions - that clearly the course is too advanced for me (specifically, starting from the third week into the SR course).

Of course, I am by no means stating that my SR professor is a bad professor. Rather, he is an incredibly smart and great professor - but, perhaps he's more suitable for a more advanced and educated audience in my opinion.
To put it shortly, I honestly wouldn't mind withdrawing from the course if I had that option readily available. However, considering that I am studying abroad as an exchange student for a year, policies state that withdrawing from a course at this point is not allowed. And, even if it were and I did so, I would consequently be forced to graduate a year later at my home university and, in turn, lose my full scholarship as a result. Needless to say, if I continue with the course and fail it, I also lose my full scholarship as well.

Therefore, the dilemma is as follows:
I - somehow - forcefully withdraw the course and lose my scholarship or continue with the course and lose my scholarship as well provided that I fail it.

Under this circumstance, I decided that still trying my best with this course and attempt to pass it would be my only best option to potentially keep my scholarships and graduate on time.

Therefore, since I have joined the Physics Forums, I really appreciate the science advisors and experts alike who have been providing me their assistance with my problems on SR. Even though I may seem really clueless at times, I nevertheless really appreciate everybody's help. Thank you!

You probably need to speak to him. This problem seems to me at best a humorous diversion.

There are a number of difficult concepts in SR that I would expect your course to cover:

Time Dilation, Length Contraction, Relativity of Simultaneity
Lorentz Transformation
Spacetime Diagrams
Four-vectors
Energy-Momentum

Have you covered all that?

Are you using a textbook?

 

[Athenian]

@PeroK

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.

Admittingly, from this point and on, I am confused on how to proceed with the solution process. And, indeed, as you have already caught on, this problem is - more likely than not - too difficult for me.

However, I do know how to complete differentiations. But, there seems to be so much going on with the provided equation I am not sure where to start ...

 

[PeroK]

@PeroK

Admittingly, from this point and on, I am confused on how to proceed with the solution process. And, indeed, as you have already caught on, this problem is - more likely than not - too difficult for me.

However, I do know how to complete differentiations. But, there seems to be so much going on with the provided equation I am not sure where to start ...

I wouldn't waste any more time on this question if I were you.

 

[Athenian]

Time Dilation, Length Contraction, Relativity of Simultaneity
Lorentz Transformation
Spacetime Diagrams
Four-vectors
Energy-Momentum

While I have used the equations of time dilation and length contraction to solve some of my questions before, it has never been taught in detail during class. At best, it was very briefly mentioned and was assumed to be basic knowledge for the students in the class. Beyond that, we have gone through all the topics you've mentioned above.

 

[PeroK]

While I have used the equations of time dilation and length contraction to solve some of my questions before, it has never been taught in detail during class. At best, it was very briefly mentioned and was assumed to be basic knowledge for the students in the class. Beyond that, we have gone through all the topics you've mentioned above.

It seems to me you've got yourself on a non-introductory course of some sort. I think you need to raise this issue.

 

[Athenian]

I wouldn't waste any more time on this question if I were you.

Under normal circumstances, that would be indeed a wise decision to make. And, for now, it would be perhaps best to move on to another question. However, if not for the fact that my entire course grade revolves around eight homework problems, I would not be trying this hard to get whatever assistance and help I could get to better understand the material as well as the problem.

Regardless, thank you for your assistance thus far and I really appreciate it!

 

[Athenian]

It seems to me you've got yourself on a non-introductory course of some sort. I think you need to raise this issue.

I'll see what I can do. However, this is indeed an "introductory" course for SR. Beyond that, this is also the only SR course in the university provided in English. And, as I need the credit for it, my home university asked me to go ahead taking it assuming the course is an "equivalent" course back at home. Anyway, thank you for the suggestion and I'll see what I am able to accomplish ...

 

[Orodruin]

This problem seems to me at best a humorous diversion.

I do not think hoodwinking students into believing that this sort of computation actually makes any sort of sense can be considered humorous at any level.

Admittingly, from this point and on, I am confused on how to proceed with the solution process. And, indeed, as you have already caught on, this problem is - more likely than not - too difficult for me.

It is clearly also too difficult for whoever posed the problem as they do not seem to realize the issues involved with posing it in the first place. Now, I am usually not very prone to dismissing other teachers out of hand so I still want to give your professor some benefit of a doubt, but there are serious issues in the formulation of the problem - including that the actual definition of the gamma factor itself lends itself to some interpretation and quite intricate details from GR that are not at all apparent if you just try an SR approach.

 

[Ibix]

Regardless, I will try my best and see what I can do with the provided equations. Of course, if you could provide additional assistance in helping me understand how I could use the following equations to "find γ as a function of r", I would sincerely appreciate it.

I very much doubt that you can get the answer the prof wants using the correct approach, sadly. As noted by others, the method you are being told to use won't give you correct answers - most obviously where $r\leq R_S$ and Newtonian gravity is laughably incorrect.

In terms of passing the course, PeroK's approach seems best. He's pretty much solved the RHS for you. Can you differentiate $\gamma$ with respect to time?

 

[Athenian]

According to a tutor that I managed to painstakingly find, he provided and taught me the following solution process for this question. I was wondering what do the science advisors here think about the solution in relation to the apparently faulty question. Thank you!

Solution:
$$\frac{\vec{v}}{c^2} \cdot \vec{F} = -\frac{\vec{v}}{c} \cdot \frac{GMm}{r^2} \hat{e}_r$$

Considering that $\vec{v} = \dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_\theta$, we get by substituting for $\vec{v}$ in the RHS $\longrightarrow$

$$\frac{\vec{v}}{c^2} \cdot \vec{F} = -\frac{GMm}{cr^2} (\dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_\theta) \cdot \hat{e}_r$$
$$\Rightarrow -\frac{GMm}{cr^2} (\dot{r} \hat{e}_r \cdot \hat{e}_r + r \dot{\theta} \hat{e}_\theta \cdot \hat{e}_r)$$

And, since $\hat{e}_r \cdot \hat{e}_r = 1$ and $\hat{e}_r \cdot \hat{e}_\theta = 0$, we get the following:
$$\Rightarrow -\frac{GMm}{cr^2} \dot{r}$$
$$\Rightarrow \frac{d}{dt} \Big( \frac{GMm}{cr} \Big)$$

Since $\frac{\vec{v}}{c^2} \cdot \vec{F} = \frac{d}{dt} (m \gamma c)$, we can continue by doing the following:

$$\frac{d}{dt} (m \gamma c) = \frac{d}{dt} \Big( \frac{GMm}{cr} \Big)$$

which yields,

$$m\gamma c = \frac{GMm}{cr} + constant$$
Or simply,

$$\gamma (r) = \frac{GM}{c^2} + constant$$

And that would be the solution. What does everybody think about it?

Once again, thank you everybody for your supportive and kind assistance throughout my many attempts in solving this question!

 

[Orodruin]

I continue to state that using Newtonian gravity in SR is inconsistent. It is particularly unfortunate as you could actually make a reasonable problem by replacing gravity by the motion of a test charge in a static electric field.

 

[Athenian]

I think everybody on the forum could agree that the question was faultily created from the beginning. And, as you've stated, the problem could have become a reasonable one by simply "replacing gravity by the motion of a test charge in a static electric field".

However, in terms of getting what the professor wants to see on the paper and ultimately passing the course, I think my solution should do the job. Well, I hope it does. What does everybody think about it?

 

[PeroK]

I think everybody on the forum could agree that the question was faultily created from the beginning. And, as you've stated, the problem could have become a reasonable one by simply "replacing gravity by the motion of a test charge in a static electric field".

However, in terms of getting what the professor wants to see on the paper and ultimately passing the course, I think my solution should do the job. Well, I hope it does. What does everybody think about it?

I must confess that I didn't notice that you only need the energy equation. That is a shortcut. You don't even need the momentum equation!

 

[Athenian]

I must confess that I didn't notice that you only need the energy equation. That is a shortcut. You don't even need the momentum equation!

Thank you! I'm glad it just worked out in the end. :)

 

[Orodruin]

I think it is very unfortunate if you need to do something that is wrong in order to pass a course. I would replace the gravitational force with an electromagnetic force with and state this in the beginning, but that is just me. It also unfortunately depends on how you think your professor would react.

Now, to the actual approach to this in GR, the problem is not well formulated because it is not clear what would be meant by $\gamma$. There are a few possibilities that include (but I probably one can think of others too):

1. The $t$-component of the 4-velocity of the particle, i.e., $\dot t = dt/d\tau$ where $\tau$ is the proper time and $t$ the time coordinate.
2. The $\gamma$ factor of the particle as measured by a static observer measuring its speed as it passes by (call this $\gamma_o$).

These two will give different final results. I would tend to favour number 2 here as it is more physical.

In the Schwarzschild solution (i.e., the spherically symmetric vacuum solution outside a spherically symmetric mass distribution), there is a constant of motion that is $(1-2GM/(c^2r)) \dot t = \gamma_\infty$, where $\gamma_\infty$ would be the $\gamma$ factor at infinity. This would also mean that, for $GM/(c^2 r) \ll 1$, you would have
$$
\dot t \simeq \gamma_\infty \left( 1 + \frac{2GM}{c^2 r}\right),
$$
which differs from the result of your problem by a factor $2\gamma_\infty$. However, as I said, my preferred interpretation would be 2, which would also require taking into account the gravitational time dilation of the static observer, leading to
$$
\gamma_o = \dot t \left( 1 - \frac{2GM}{c^2 r}\right)^{-1/2} = \gamma_\infty \left( 1 - \frac{2GM}{c^2 r}\right)^{-3/2} \simeq \gamma_\infty \left( 1 + \frac{3GM}{c^2 r}\right),
$$
which differs to the result of your problem by a factor $3\gamma_\infty$. (Assuming I did the math correctly here, it was kind of hastily between other tasks ...)

 

[Athenian]

I think it is very unfortunate if you need to do something that is wrong in order to pass a course. I would replace the gravitational force with an electromagnetic force with and state this in the beginning, but that is just me. It also unfortunately depends on how you think your professor would react.

Now, to the actual approach to this in GR, the problem is not well formulated because it is not clear what would be meant by $\gamma$. There are a few possibilities that include (but I probably one can think of others too):

1. The $t$-component of the 4-velocity of the particle, i.e., $\dot t = dt/d\tau$ where $\tau$ is the proper time and $t$ the time coordinate.
2. The $\gamma$ factor of the particle as measured by a static observer measuring its speed as it passes by (call this $\gamma_o$).

These two will give different final results. I would tend to favour number 2 here as it is more physical.

In the Schwarzschild solution (i.e., the spherically symmetric vacuum solution outside a spherically symmetric mass distribution), there is a constant of motion that is $(1-2GM/(c^2r)) \dot t = \gamma_\infty$, where $\gamma_\infty$ would be the $\gamma$ factor at infinity. This would also mean that, for $GM/(c^2 r) \ll 1$, you would have
$$
\dot t \simeq \gamma_\infty \left( 1 + \frac{2GM}{c^2 r}\right),
$$
which differs from the result of your problem by a factor $2\gamma_\infty$. However, as I said, my preferred interpretation would be 2, which would also require taking into account the gravitational time dilation of the static observer, leading to
$$
\gamma_o = \dot t \left( 1 - \frac{2GM}{c^2 r}\right)^{-1/2} = \gamma_\infty \left( 1 - \frac{2GM}{c^2 r}\right)^{-3/2} \simeq \gamma_\infty \left( 1 + \frac{3GM}{c^2 r}\right),
$$
which differs to the result of your problem by a factor $3\gamma_\infty$. (Assuming I did the math correctly here, it was kind of hastily between other tasks ...)

Wow, thank you for going through the calculation process here. I definitely learned a lot through this. Beyond that, I do agree that it is quite unfortunate that I have to answer something incorrectly to pass the course. While I did entertain the thought of replacing the gravitational force with the electromagnetic force upon your suggestion, I know with certainty that would guarantee me a failing mark on that specific problem. Therefore, even if the structure of the question is wrong to begin with, it is best I follow in accordance to the instructor's guidelines.

Beyond all that, though, I do sincerely appreciate the work you've showed here to help me gain a better understanding of what a structured problem and solution should look like here instead. Thank you very much!

 

[Athenian]

$\gamma (r) = \frac{GM}{c^2} + constant$

Ah, my apologies. I accidentally missed the $r$. The answer's supposed to be:
$$\gamma (r) = \frac{GM}{rc^2} + constant$$

Of course, please let me know if I'm mistaken. Thank you!

 

[Orodruin]

I do agree that it is quite unfortunate that I have to answer something incorrectly to pass the course.

Actually, it is beyond unfortunate as it puts you in a bit of moral dilemma also in regards to future students. If you just go along and hand in the expected solution then there will be no change for the next course offering and a new batch of students will face the same errors. On the other hand, it is not necessarily obvious that alerting your professor to this will have an impact either. You might want to consider your options based on the approachability of your teachers. I can understand that you as a student do not feel like you have the authority to challenge your professor, but you can always send your teachers here to discuss the issue with us. There is quite a number of people here that are quite well versed with regards to relativity.