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July 11

Relativistic Doppler shift

Hi!

I am working on User:YohanN7/Relativistic Doppler effect. The purpose is to get it right, shorten it down and then to include it in another article (velocity-addition formula).

But, alas, I am stuck. In order to get the supposedly correct answer, given in Relativistic Doppler effect#Systematic derivation for inertial observers without reference, I need to introduce an ad hoc minus sign. I have noted clearly where I do it, and what "good" it does.

A similar derivation exists in Relativistic Doppler effect#Systematic derivation for inertial observers. There, they also play tricks with signs when transforming velocities. I don't really understand it.

There is one difference in notation. I take all velocities relative to the positive direction of the x-axis. The problem I have does not go away by switching from $L = T (s + V)$ to $L = T (s - V)$ , hence assuming an approaching emitter has positive velocity. I still need the minus sign magic (and then get exactly the formula in Relativistic Doppler effect#Systematic derivation for inertial observers; with my convention there is a difference in sign as there should be).

I have been starring myself blind at this. There is probably a trivial explanation that I simply can't see. Maybe I set up the problem incorrectly all together. Help appreciated, either with pointing out my mistake, or pointing me to a reference where this particular version of Doppler shift is actually treated. YohanN7 (talk) 18:01, 11 July 2015 (UTC)[reply]

@YohanN7: Can you clarify what you're trying to calculate? A wave speed s < c must be given with respect to some medium, so the result will depend on s and two relative velocities (e.g. the velocities of emitter and receiver relative to the medium), but there's only one relative velocity in the final formula. I think that formula is for the case that the receiver is at rest relative to the medium. But the text implies that the speed s' appearing in the formula is the wave speed relative to the emitter. The speed in the formula should be relative to the medium/receiver. That could be the cause of the sign error. -- BenRG (talk) 07:49, 14 July 2015 (UTC)[reply]

Thanks for the reply. Yes, I realize that a medium is definitely physically involved for this to occur, and at first sight, velocities relative to this medium should be used. However, in Relativistic Doppler effect#Systematic derivation for inertial observers, they calculate the same thing as I am after, and they also do not even mention the medium. They point at which they play around with minus-signs is equally magical to me.

In my notation, there are three velocities involved, all measured in the positive

x

-direction,

V, s', s

. These are emitter velocity as seen in the unprimed frame, velocity of (a particular) wave front in the

x

-direction as seen in primed system, and velocity of the same wave front in the

x

-direction as seen in the unprimed system. The corresponding quantities in the article are

v, u', u

. They also mix in speed for the named quantities (by def ≥ 0).

My "playaround":

-s={\frac {s'+V}{1+{s'V \over c^{2}}}}.

Theirs:

-u={\frac {-u^{\prime }+v}{1+(-u^{\prime }){\frac {v}{c^{2}}}}}.

The non-substantial difference is that they here have the

u, u'

as speeds > 0, while my

s, s'

are velocity components allowed to take on negative values. It is the minus sign on the left that I am forced to introduce that bothers me.

There is no mention of media and velocity relative to it, and I think it isn't needed in this case. You could visualize the same set up where the emitter is shooting bullets with a relativistic machine gun. What I mean is that it need not be waves, the treatment is in that sense more general. We could count bullets instead of wave fronts. YohanN7 (talk) 11:35, 14 July 2015 (UTC)[reply]

Perhaps it is the case that

s={\frac {s'+V}{1+{s'V \over c^{2}}}}.

is entirely correct (this is, after all, what the velocity addition formula gives), and the error lies in that I should really use

- s

in my original formula from the set up. I'd be entirely happy if this is so. YohanN7 (talk) 11:59, 14 July 2015 (UTC)[reply]

My set up includes

L=sT+VT=(s+V)T,

where

L

is the observed wave length (distance between bullets) in the unprimed system. Assuming negative

s

, so that the wave crests are approaching,

sT

gives a negative contribution, resulting in a shorter wave length for a receding emitter. This defies common sense, and I believe that this is the explanation. The set up was copied from the corresponding scenario for light waves, where the speed of light is involved. Embarrassing, but I was looking in the wrong place

. @BenRG:, I'd be grateful if you could either refute or confirm this observation, since it is intended for an article and I have no reference at hand (neither does the Relativistic Doppler shift article treating the same thing). I have updated my draft with "magic" and "no magic" in parallel. YohanN7 (talk) 13:15, 14 July 2015 (UTC)[reply]

By the way, a nicer way of writing the colinear redshift in a medium is

{\frac {\nu _{e}}{\nu _{r}}}={\sqrt {\frac {c^{2}/v_{wr}^{2}-1}{c^{2}/v_{we}^{2}-1}}}

, where v_wr is the relative speed of the wave and receiver and v_we is the relative speed of the wave and emitter. A simpler and manifestly covariant version of that is

{\frac {\nu _{e}}{\nu _{r}}}={\frac {\sinh \alpha _{we}}{\sinh \alpha _{wr}}}

, where α is relative rapidity. It's easy to see why that's correct, too: consider the problem in the rest frame of the wave, where the emitter passes by and leaves two stationary wave fronts and the receiver later passes by and collects them, and remember (or work out) that sinh α is the coordinate distance you travel per unit proper time if your rapidity is α. For general (not necessarily colinear) motion, still in a medium,

{\frac {\nu _{e}}{\nu _{r}}}={\frac {\gamma _{me}-\gamma _{mw}\gamma _{we}}{\gamma _{mr}-\gamma _{mw}\gamma _{wr}}}

(where m stands for medium). For general motion and a light signal in vacuum,

{\frac {\nu _{e}}{\nu _{r}}}={\frac {\mathbf {x} \cdot \mathbf {v} _{e}}{\mathbf {x} \cdot \mathbf {v} _{r}}}

where v_e and v_r are the four-velocities of the emitter and receiver and x is the (lightlike) direction of the light propagation. A completely general formula combining all of the above is

{\frac {\nu _{e}}{\nu _{r}}}={\frac {\mathbf {x} ^{2}(\mathbf {v} _{m}\cdot \mathbf {v} _{e})-(\mathbf {x} \cdot \mathbf {v} _{m})(\mathbf {x} \cdot \mathbf {v} _{e})}{\mathbf {x} ^{2}(\mathbf {v} _{m}\cdot \mathbf {v} _{r})-(\mathbf {x} \cdot \mathbf {v} _{m})(\mathbf {x} \cdot \mathbf {v} _{r})}}

. It would be nice to have some of those in Wikipedia, but I don't have a source for them. I only ever see the ugly coordinate versions in textbooks. -- BenRG (talk) 08:51, 14 July 2015 (UTC)[reply]

Some basic things

How can I solve problems like "What's the smallest natural number that, when divided with a, b and c, gives the remainder d"? I wasn't taught this at school but it is in my homework. --Biolongvistul (talk) 18:34, 11 July 2015 (UTC)[reply]

d+LCM(a,b,c). Bo Jacoby (talk) 20:25, 11 July 2015 (UTC).[reply]

Hmm, the question as phrased is not entirely clear. Bo seems to be reading it as "find n such that the remainder when you divide n by a is d, when you divide n by b is also d, and when you divide n by c is also d". Is that what you mean? If so, then I think Bo's answer is correct, granted that d is smaller than the smallest of a, b, and c. If d is not smaller than that, then there is no answer.

More generally, see Chinese remainder theorem. --Trovatore (talk) 20:51, 11 July 2015 (UTC)[reply]

Yes, that is what I meant. Thank you for your answer. --Biolongvistul (talk) 13:44, 12 July 2015 (UTC)[reply]

If they meant find the smallest x such that ^x/_(abc) leaves a remainder of d, then the solution is just (abc)+d. StuRat (talk) 21:02, 11 July 2015 (UTC)[reply]

Actually, now that I think about it, the solution in both cases is d. --Trovatore (talk) 21:07, 11 July 2015 (UTC)[reply]

Would that be a likely homework question? I wonder whether, as Trovatore suggests, the Chinese remainder theorem is implied and, if so, the method of successive substitution might help. (My homework was never as difficult as that though!). Thincat (talk) 21:59, 11 July 2015 (UTC)[reply]

Even if it is homework, they are asking how to solve that class of problems, not a specific case. Had they given values to a, b, c, and d and asked for a numeric answer, then a "Do your own homework" template would be in order. StuRat (talk) 23:13, 11 July 2015 (UTC)[reply]

Not sure why the answer isn't d. d<d+LCM(a,b,c) and d divided by a, b, or c has a remainder of d. Unless d is isn't smaller than a, b, or c in which case there is no answer; so I'm taking d<a,b,c as implicit. As for the general class of problem, not sure what that is with only one example, but Chinese Remainder Theorem sounds like a good guess, all remainders equal being a simple case. --RDBury (talk) 01:33, 12 July 2015 (UTC)[reply]

Yes, the trivial answer is d, but that is not what the professor had in mind, so the bright student answers d+lcm(a,b,c). Bo Jacoby (talk) 01:45, 12 July 2015 (UTC).[reply]

Yes, you have to assume that the unknown number must be divided at least once by a, b, and c to leave a remainder of d, to make the problem interesting. StuRat (talk) 01:51, 12 July 2015 (UTC)[reply]

"Row upstream at 2 knot and back downstream at 3 knot. Average speed?" I feared the answer "2.5 knot" and hoped for the answer "2.4 knot", but I got the answer "0 knot - you ended where you started". Bo Jacoby (talk) 15:16, 12 July 2015 (UTC).[reply]

That's the difference between average speed and average velocity, at least way back when I took high school physics. But, as often happens when you talk about velocity, I'm afraid we're getting off on a tangent. --RDBury (talk) 15:45, 12 July 2015 (UTC)[reply]