Particle Deceleration in Relativistic Jets?

[PeterDonis]

I thought "for the sake of argument" you were earlier willing to discuss a close approximation of a uniform 1m/s^2 region.

You misunderstood my intent. It was to emphasize the fact that you were making an assumption instead of actually looking at the applicable numbers. The topic of the thread is relativistic jets and your question about deceleration. In order to answer that question with reference to the jets you asked about, we need to be working with numbers that are applicable to those jets.

I am asking if my calculation of the amount of energy lost by a relativistic electron in such a region is approximately accurate.

And I've said that the point made by @kimbyd, that the "deceleration" actually applies to momentum, not velocity, is valid. Does your calculation take that point into account?

 

[metastable]

Does your calculation take that point into account?

I believe it factors the momentum since the calculation is based on total electron kinetic energy with upward 1m/s in a roughly 1m/s^2 field, not velocity, but I am not 100% confident, which is why I am asking if the answer I calculated is correct. The topic is not about those particular jets but relativistic jets in general. I want to know how to calculate deceleration in relativistic jets in general, based on the mass of the black hole, the distance of the particle and the particle’s upward velocity/momentum/energy.

 

[metastable]

And I've said that the point made by @kimbyd, that the "deceleration" actually applies to momentum, not velocity, is valid. Does your calculation take that point into account?

I used this formula in an attempt to calculate the energy lost by the relativistic electron in nearly uniform 1m/s^2 field by using V=1m/s:

which if correct implies:

The upward V=299792457.9999608591483m/s electron (1000gev to hovering observer) loses approximately 0.0000000000028428ev in one second to the hovering observer in a gravitational field which closely approximates uniform 1m/s^2...

...though I am not 100% confident. Did I use the right formula?

 

[Ibix]

Argument: a non-relativistic target in a uniform weak gravitational field absorbs a relativistic electron, thereby increasing the target's mass by $\gamma m_e$. The target is slowly lowered a distance $\Delta h$ in a uniform gravitational field before emitting the electron again upwards at the same $\gamma$ it had before. By the time the electron has climbed $\Delta h$ it had better have lost energy $\gamma m_eg\Delta h$, or else I can exploit this to get free energy. Therefore, in a uniform weak gravitational field, $\gamma m_eg\Delta h$ ought to be a decent approximation for the energy lost by a relativistic electron.

Using a gamma factor of 2×10⁶ (giving $\gamma m_e$ of about 1000GeV), mass of 9×10^-31kg, g of 1ms^-2 and a height change of 3×10⁸m gives you 54×10^-17J, or about 3.4keV. Assuming both that my approximation and my arithmetic are correct.

 

[Ibix]

Incidentally, right or wrong, that's how to lay out maths. Note that I've explained what all the symbols mean (except standard ones like $\gamma$ and $m_e$), I've used LaTeX for formulae, and I have not expressed any number to more than a couple of significant figures.

Compare that to your gigantic wall of near-identical sixteen significant figure numbers with little to no explanation of why you were combining numbers the way you were doing. Which one is easier to understand? Which one will people be willing to put in the time (that they are giving for free, for the fun of it) to read?

 

[metastable]

Argument: a non-relativistic target in a uniform weak gravitational field absorbs a relativistic electron, thereby increasing the target's mass by γmeγme\gamma m_e. The target is slowly lowered a distance ΔhΔh\Delta h in a uniform gravitational field before emitting the electron again upwards at the same γγ\gamma it had before. By the time the electron has climbed ΔhΔh\Delta h it had better have lost energy γmegΔhγmegΔh\gamma m_eg\Delta h, or else I can exploit this to get free energy. Therefore, in a uniform weak gravitational field, γmegΔhγmegΔh\gamma m_eg\Delta h ought to be a decent approximation for the energy lost by a relativistic electron.

Using a gamma factor of 2×106 (giving γmeγme\gamma m_e of about 1000GeV), mass of 9×10-31kg, g of 1ms-2 and a height change of 3×108m gives you 54×10-17J, or about 3.4keV. Assuming both that my approximation and my arithmetic are correct.

Compare that to your gigantic wall of near-identical sixteen significant figure numbers with little to no explanation of why you were combining numbers the way you were doing. Which one is easier to understand? Which one will people be willing to put in the time (that they are giving for free, for the fun of it) to read?

Thank you for your thoughtful analysis. My problem solving methods are a bit unconventional in a physics setting... but they evolved from a programming / coding background and in this setting we don't usually convert to scientific notation or eliminate significant decimal places. This is a rather complex vehicle performance chart I made previously, note that despite retaining all decimals it's still quite readable when graphed:

https://forum.esk8.news/uploads/default/original/2X/9/9eb5f2adb5179883cb6b17982dbde99c568499ad.jpeg

I had hoped to apply the same programming/graphing methods I use when solving other complex scenarios to the topic at hand, particle deceleration in relativistic jets.

 

[PeterDonis]

Did I use the right formula?

As far as I can tell, you are using the SR formula for energy as a function of velocity. I don't see how that's relevant to energy change with height change in a gravitational field.

@Ibix gave you an excellent example of how to show your work for a problem like this. Unfortunately, your very next post, #36, is worse, not better.

My problem solving methods are a bit unconventional in a physics setting

Your problem solving methods are not the issue. Your apparent inability or unwillingness to present your results in a way others can comprehend without inordinate effort is.

it's still quite readable when graphed

You must be joking.

 

[Ibix]

My problem solving methods are a bit unconventional in a physics setting... but they evolved from a programming / coding background

I'm not sure I know a physicist who can't program to some level or other in two or three languages. The problem isn't to do with programming. It's that you don't explain what you are doing and don't define your terms. And in response to my criticism you provide a screenshot with so many numbers I cannot even make out the digits, with no explanation of what they are, how you got them, or why you think they are relevant to the problem at hand. If you code like that then you produce the kind of unmaintainable uncommented mess that I'm currently spending my working life rewriting because nobody understands or trusts it.

I'm sure you can do arithmetic. What you don't seem to grasp is physics. So you need to state clearly what problem you are trying to solve, which formulae you are using, why you think they are relevant, and how you are combining them. Only at that point should any numbers be introduced. Using the template from the homework forums would not be a bad idea.

Note that there is one formula in my post #34, embedded in a whole paragraph of justification for why I think it's appropriate. And I use LaTeX to make it easy to read. Then I simply state the numbers I'm going to put in and the result I got. Look at any scientific paper and you will see the same. It's the reasoning that's important, and laying it out for ease of comprehension, not a blow-by-blow account of you using your calculator.

 

[Ibix]

For example, see this thread, which was me asking about something I wasn't quite sure I understood. Note that I provided a link to where I got the equations from, explained what I was trying to do, and provided what I thought was the correct answer and how I got to it. Peter was immediately able to point out where I had gone wrong, and George offered more help when I stated a problem I was having with the numerical work. Again, there are no numbers except implicitly in the graphs (which don't appear until several posts in), just explanation of what I was doing, why, and how I got there.

 

[metastable]

You must be joking.

And in response to my criticism you provide a screenshot with so many numbers I cannot even make out the digits

For example, see this thread, which was me asking about something I wasn't quite sure I understood. Note that I provided a link to where I got the equations from, explained what I was trying to do, and provided what I thought was the correct answer and how I got to it. Peter was immediately able to point out where I had gone wrong, and George offered more help when I stated a problem I was having with the numerical work. Again, there are no numbers except implicitly in the graphs (which don't appear until several posts in), just explanation of what I was doing, why, and how I got there.

Thank you, I do appreciate the criticism, I know it's well meaning. I know it seems like I was joking about the readability of the chart since the numbers were all blurred, but that was an effect of physics forums resizing the image to a smaller size when I uploaded it, which is why I had included a direct link to the original:
https://forum.esk8.news/uploads/default/original/2X/9/9eb5f2adb5179883cb6b17982dbde99c568499ad.jpeg

 

[Vanadium 50]

Your continued refusal to take Ibix's excellent advice is puzzling.

 

[metastable]

I agree it's excellent advice.

0.0000000000028428ev in one second

@Vanadium 50 -- 2.84*10^(-12)eV in one second

about 3.4keV

So is anyone certain if either one of these answers is correct?

 

[metastable]

3.4keV

@Ibix I'm troubled by this answer (3.4keV) because by rearranging the equation below, I believe it can be shown that, ignoring gravity, 3.4keV is sufficient energy to accelerate an electron from standstill to V = 3.44*10^7 m/s, so if we look at the effects of a 1m/s^2 gravity field acting on an electron, how can a 1m/s^2 field have enough "power" to remove sufficient energy in one second, that it is equivalent to changing a stationary electron's velocity by ~3.44*10^7 m/s?:

 

[Ibix]

Because the circumstances are completely different. What is $\Delta h$ in the two cases of an electron starting from rest and an electron traveling at 0.99c?

 

[metastable]

What is $\Delta h$ in the two cases of an electron starting from rest and an electron traveling at 0.99c?

If I understand the implications of this question correctly, you are saying it takes more energy to lift an electron which is initially at rest to a hovering observer to the height that a 0.99c electron can cover in one second, than lifting to the height a 1m/s electron can travel in the same time.

What I am saying is if the same amount of force acts on both particles for the same amount of time to the hovering observer, wouldn't the same amount of energy be "removed" from either particle to another hovering observer after the same 1 second duration of time, whether the particle is initially moving upwards at 1m/s or moving upwards at 0.99c?

 

[Ibix]

Why not answer the question, instead of second guessing me?

 

[metastable]

What is ΔhΔh\Delta h in the two cases of an electron starting from rest and an electron traveling at 0.99c?

0.99 * 2.998e+8 m

Isn't the acceleration to a hovering observer implied by 1m/s^2 proportional to time not distance?

 

[Ibix]

And what's $\Delta h$ in the other case, with the electron starting stationary?

There shouldn't be a $c$ in your previous answer, by the way.

 

[metastable]

-1m

 

[metastable]

There shouldn't be a ccc in your previous answer, by the way.

Sorry it was meant as a label not a multiplication factor... fixed.

 

[metastable]

-1m

Unless I'm mistaken, If a collision with another hovering observer occurred at this point, 2.84*10^(-12)eV kinetic will be measured.

 

[Ibix]

-1m

You've missed a factor of a half here. Anyway, the main point is that one electron changes height by more than a hundred million times more than the other. Why would you expect the energy change (which you know depends on $\Delta h$) to be the same?

 

[metastable]

You've missed a factor of a half here.

Thank you, you are correct.

 

[metastable]

Anyway, the main point is that one electron changes height by more than a hundred million times more than the other. Why would you expect the energy change (which you know depends on ΔhΔh\Delta h) to be the same?

It would require outside energy to lift a stationary electron (to a hovering observer) to a height of + 0.99 * 2.998e+8 m, but no additional outside energy is necessary to "lift" an upward 0.99c electron to the same height. So I guess I don't understand why we would even look at the energy it takes to lift stationary electrons a given distance, because in the scenario we were calculating in which you gave an answer of 3.4keV, I believe the amount of additional outside energy required to lift either the upwards 0.99...c electron the distance it travels in one second, or the upwards 1m/s electron the distance it travels in one second, is precisely 0.

 

[PeterDonis]

I know it seems like I was joking about the readability of the chart since the numbers were all blurred

The numbers being blurred is not the issue. Unblurring the numbers doesn't help. There is no explanation of what these charts are intended to show, where you got the numbers, what the quantities mean, etc. Of course it all makes sense to you because you made the charts and you already know all the background information behind them. But we don't know any of that, and without it your charts are gibberish.

 

[Ibix]

So I guess I don't understand why we would even look at the energy it takes to lift stationary electrons a given distance,

I appear to have confused the scenario you were talking about in #43 with the one you were talking about in #13. It would help if you could focus on one scenario.

In either case, the point is that the work done is the force applied times the distance moved. And the distance moved in unit time is not the same for things moving at different velocities.

 

[metastable]

In either case, the point is that the work done is the force applied times the distance moved.

I'm not contradicting, just presenting the definition of force I'm looking at:

 

[PeterDonis]

just presenting the definition of force I'm looking at

Which is headed by the words "Special Relativity". Which doesn't work when gravity is present, i.e., when spacetime is curved. You cannot use SR in the scenario you are trying to analyze. This has been repeatedly pointed out to you, yet you continue to ignore it.

 

[PeterDonis]

The OP question has been addressed. Thread closed.