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Questions of absolute light speed and mass by an engineer

  1. Dec 4, 2015 #1
    Hello everybody. I am an spanish student of a bsc in Chemical Engineering, so as you can imagine, all this questions are going to be purely curiosity. First of all, sorry for any english language mistake I can commit.

    I have always been curious of the spacetime issues, and how the world really works. My intelectual formation leaded me to some consistent answers to many questions, but, inevitably, they were newtonian answers. When it comes to "understanding" relativity, my reasoning wasn't enough. Let me tell you one of my believes, whether it's correct or not, so you can have a clue of how does my mind starts to think;
    I don't understand time as a variable, I simply understand it as a way of measuring events in a space where there is no past or future, we are simply in a static reality where events take place, and we needed a tool to measure two different events. It's, without a doubt, a really useful tool. But still doesn't exist. (let me repeat this is what I think, not trying to lecture). I know this belief might be an issue when understanding relativity, but my mind is too newtonian. (I'm an engineer, so don't blame me).

    Now, let me develop a few more some questions.

    In order to make things a bit clear, I bought the famous book "A briefer history of time", of the well known Hawking. I preferred a book a few more advanced, as I can understand maths and physics in a bit higher level than a ungraduated person, but didn't know where to start, and this was a good point. Started reading, and I got stuck at page 40. I stopped reading when it mentioned the concepts of relative mass, and the absolut speed light for any observer.

    I am going to center the thread on this two concepts. Let's start with the first one. It's not going to be a "rel. mass vs rest mass" battle as it has been all this days, so don't be afraid to continue reading.

    I went to the net and started to investigate this concept of relative mass, and found the internal battle in physics for eliminate the concept of relativity mass. Quoting Taylor and Wheeler, in "Spacetime Physics", we can read;
    This last sentence cleared me the enormous doubt I had when reading the book. Hawking said that, given E=mc2, if we increase the energy of a certain object, for example with speed, as kinect energy, the mass would INCREASE given the fact that light speed is a constant. Wtf was that, my newtonian mind was frustrated. I read about the energy-mass equivalences, but still don't believe. But at least I wasn't alone, there were a bunch of scientifics denying the concept.
    I've read about that, in nuclear fision, if you sum the masses of the final atoms, you won't get the mass of the initial atom. That's hard to understand, but didn't find much information, so I simply anotated it but didn't focus on.

    So, let me ask you; how can I understand this formula? What I'm going to say may hurt you, and I am going to say it so you can teach me where am I wrong; Was Einstein bad? I mean, given that E=mc2 is true, the ONLY way to keep mass constant, when applying energy to an object, is that lightspeed decreases. And this is how I'm connecting this issue with the second I talked you two pharagraphs before;

    In the Michelson and Morley experiment, they demonstrated that the light was arriving to the earth exactly at the same speed, didn't matter if we, the earth, were moving towards or against the source of light (sun), but we were always receiving it at the SAME speed. No stories abut sums of speeds and that. Light had an ABSOLUTE speed, and doesn't matter if the observer is moving or not.
    Let me tell you this experiment completely freaks my mind. Can't understand. I honestly don't believe the calculations were accurate enough. (Let me remind you that all I say is intented so you can explain me a bit about this). How can you measure a change on a photon, going at ~300.000 km/s, with the movement of earth, at ~30 km/s? If the results on the measures were the same when moving towards and opposite the sun, is beacuse they weren't accurate enough, and the change was SO small they didn't even have tools to measure it.

    Imagine you and a friend of yours, both in different spaceships. You are going to stay static on space, right next to the sun (let's keep the temperature and disintegrating issues appart). Your friend is going to be next to you, also static for a moment. Imagine you can see a photon leaving the sun. You fix your eyes onto it and stay there, next to the sun, and your friend burns the engine and starts to move at 10.000 km/s in the direction of the photon. The absolute lightspeed concept says that you would see the photon moving at 300.000 km/s, and that your friend, moving towards the photon, would ALSO see it moving at 300.000 km/s. Excuse me but I truly can't understand this. How is this photon going to have different speeds in a single universe?
    If the photon is moving at 300.000 km/s for you, your friend should see it at 290.000 km/s, as he's moving towards at 10k km/s. I know this is purely newtonian but wanted to make it clear. How can I believe this concept of absolute speed?

    Summing up, I don't believe in mass changes, I don't believe in absolute light speed, I don't believe in time, and I don't believe in speedlight being the maximum speed we can get. Help me before my mind explodes, please.

    Thank you for your time.

    EDIT: Forgot to ask something. There was an experiment were there were one plane which carried an atomic clock, perfectly syncronized with another one which was on earth. The plane flied for an amount of time, and then, comparing the clocks, there was a desyncronization (not sure how to write that on english). The time passed more slowly to the one on the plane. Please, how can a physical object, manufactured to work in a single way, give a change on the resulting time?? Physical objects CAN'T demonstrate a change on the time. A clock will measure a second exactly the same way, flying at lightspeed or staying without moving.
    Please I really need you to make this clear to me, because Interestellar is my favourite movie, this argument of mine is freaking with this film in my mind, and I really want to keep interestellar as my favourite movie (lol)
     
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  3. Dec 4, 2015 #2

    A.T.

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    Use the search function. All of this has been discussed here many times.
     
  4. Dec 4, 2015 #3
    I have been searching a bit, but can't find a response to the absolute light speed issue. Sorry if I simply couldn't find it

    EDIT;
    Just find a clue on the first page, by user HallsofIvy ;

    If object B is moving at speed u relative to observer A and object C is moving at speed v relative to object B then A will observe object C moving at speed [tex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/tex] relative to himself. If object C is a photon, so that v= c, then that reduces to [tex]\frac{u+ c}{1+ \frac{uc}{c^2}}= \frac{u+ c}{1+ \frac{u}{c}}= \frac{c(u+ c)}{u+ c}= c[/tex]. Light moves at speed c relative to any observer.

    I am going to give it a deep look after I study for an exam I have nearby; want to make this clear for my mind. Would you please explain a little bit more, how did you get the first equation, of the velocity of the object C for the observer A? I'm sure it's basic physics, but it's been a long time since I don't use this things, since the first year of the bachelor. Thanks
     
    Last edited: Dec 4, 2015
  5. Dec 4, 2015 #4

    Dale

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    Hi nanodean, welcome to PF!

    The formula ##E=mc^2## is correct, but it only applies when the momentum, ##p=0##. For a moving system the equation is ##E^2/c^2=m^2 c^2+p^2##. Hopefully it is easy for you to see that the general formula reduces to the more famous one for ##p=0##. In this formula, c remains constant regardless of p.
     
  6. Dec 4, 2015 #5
    Thank you so much for your help. Asuming p=m·v, and that we are talking about a non massless object, E=mc2 is valid when v=0, so we get the rest energy. The other equation you gave me perfectly satisfies my question about this strange mass increasing it seemed to be on the original equation. I guess that with this formula we can get the energy of an object when its speed is NOT zero.

    Thank you again, this clarified me a lot
     
  7. Dec 4, 2015 #6

    Nugatory

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    ##E=mc^2## gives you the energy of an object that is at rest relative to you. If you apply energy to it and accelerate it, then its total energy includes some kinetic energy and you use ##E^2=(mc^2)^2+(pc)^2## where ##p## is the momentum. Note that the latter reduces to the former when the object is at rest so ##p## is zero.

    They used an interferometer (although we don't generally recommend issuing wikipedia as a source, the article on the "Michelson Morley interferometer is OK). This device is capable of detecting differences in travel time that are less than the time for one wavelength to come by, and one part in 104 is easy.
     
  8. Dec 4, 2015 #7

    Dale

    Staff: Mentor

    You should definitely start here:
    http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

    The original MMX was not super-sensitive, but certainly better than 1 part in 10^4 which is actually not that challenging as you seem to think. Modern experiments have brought that limit down to 1 part in 10^18.

    The first thing is to forget your beliefs. They simply are not relevant. The universe does not care about your beliefs, and frankly your beliefs are uninformed.

    What you should focus on initially instead is understanding. Understand how relativity works and what experimental evidence says. After understanding and information then belief will likely follow.
     
  9. Dec 4, 2015 #8

    Nugatory

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    Yes, a properly constructed clock will tick exactly once per second - that's what a clock does. It measures the seconds going by the same way that the odometer of an automobile measures the kilometers passing under the wheels. If two automobiles start at the same point, and travel by different routes to the same destination, and we find that one odometer reading has increased by ten kilometers and the other by twelve kilometers we conclude - because they both measure distance the same way no matter what the speed! - that one route was two kilometers shorter than the other. In the Hafele-Keating experiment we sent two clocks on different routes between the same two points in spacetime (plane takes off and plane lands); they each ticked once per second just as the odometer of a car ticks once every kilometer on a path through space; they ticked a different number of times along their paths through spacetime because less time passed on one path than the other.
     
  10. Dec 4, 2015 #9
    I am reading all of you guys, this evening (europe timezone) I'll look your answers with a bit deeper read so I can reply properly. My mind should be on the exam but it's on this (lol)
     
  11. Dec 4, 2015 #10
    Okay, I was wrong for the assumption that they simply weren't accurate enough - It was simply a way to justify my own believes. Now, seeing that the experimental part is getting more sense, I am understanding a bit more.

    Thanks for the link. It's dense but I'll be giving it a look when I have freetime gaps. Did a quicksearch for the re-made MMX and saw the precision.

    Don't take that last part of my post totally serious, it was simply a way to describe how was my mind rejecting the relativity, maybe that made you more clear from where should I start learning given the things I was stuck in. Don't know if it helped in that way, but you definitely helped me in key issues.

    I got the idea. So, if that plane simply took a way where time went slowly, we are saying that time dilates with speed. As it is also known, it also dilates in presence of a gravity field. So we can say:

    -The more gravitational effects, the more dilation of time (which means, the closer you are to a big mass, the slower time goes for you)
    -The more speed, the more dilation of time (which means, the faster you go in a certain direction, the slower times goes for you)

    Are those two statements correct?
    Interestellar is getting stable therefore I am getting happy. I know it's not the most correct film in terms of scientific issues, but just want to keep my 9 in filmaffinity for it (lol).

    Now I'm just missing a concept. I posted in this thread an equation I saw by a guy I quoted. I contacted him for the origin of it, but maybe you can also help me. I guess this is now simply dynamics, but my mind has too far the courses of Physics I took on the first years of the engineering bachelor.

    If object B is moving at speed u relative to observer A and object C is moving at speed v relative to object B then A will observe object C moving at speed [tex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/tex]

    So; how did he got the equation of the velocity of C by the observer A?

    With this, I think I can close my load of questions. I am pretty satisfied with the concepts I learned. Thank you all.
     
  12. Dec 4, 2015 #11

    Nugatory

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    If you google for "relativistic velocity addition derivation" you'll find some good derivations.

    However, if you are serious about understanding relativity, you'll be best served by working through things in a more disciplined order:
    1) Understand how the uniform speed of light leads to relativity of simultaneity (google for "Einstein simultaneity train"). Until you have relativity of simultaneity down, nothing is going to make sense.
    2) Understand how the uniform speed of light leads to the Lorentz transformations, because everything builds on them. Again, there are many good derivations on the web; and although generally modern textbooks are better than the original historical writings, in this particular case the algebra-only derivation in the appendix of Einstein's popularization "Relativity; The special and general theory" is actually pretty good.
    3) Use #1 and #2 to derive the time dilation and length contraction formulas for yourself.
    4) Work through the two elementary "paradoxes" of special relativity: Bug-rivet and pole-barn. Google works.
    5) Learn about space-time intervals and the Minkowski stuff. You could do much worse than Taylor and Wheeler's "Spacetime Physics" for this part, and actually it's a good guide for the entire process.
    6) Work through the twin paradox, starting with http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
     
  13. Dec 4, 2015 #12
    Thank you for both advices. I'll answer the second first; I am serious about getting a concept and being able to understand relativity at a "pedestrian" level, but unfortunately I don't have the time nor the mathemathical handling to dedicate, and I consider this a curiosity hobby. However, I will still give a look at the steps you gave me as soon as I have freetime gaps.

    Secondly, you iluminated me with the "relativistic velocity addition derivation". I've developed it, taking for granted the Lorentz Transformations, which I'll give a deeper look later, and ended up with this two equations;

    [tex]u=\frac{u'+ v}{1+ \frac{u'v}{c^2}}[/tex] [tex]u'=\frac{u- v}{1- \frac{uv}{c^2}}[/tex]

    Where:
    u is the velocity of a particle by observer 1 (static)
    u' is the velocity of the same particle by observer 2 (moving)
    v is the speed at which observer 2 is moving
    c is the speed of light

    If we make this particle a photon, we would have;

    For the static observer
    [tex]u=\frac{u'+ c}{1+ \frac{u'c}{c^2}}= \frac{u'+ c}{1+ \frac{u'}{c}}= \frac{c(u'+ c)}{u'+ c}= c[/tex]

    For the moving observer
    [tex]u'=\frac{u- c}{1- \frac{uc}{c^2}}= \frac{u- c}{1- \frac{u}{c}}= \frac{c(u- c)}{c- u}= \frac{c(u- c)}{-(u- c)}= -c[/tex]

    What am I doing wrong that I get a negative speed on the moving observer?
     
  14. Dec 4, 2015 #13

    Nugatory

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    Those are velocities, not speeds, so they have a direction. The negative velocities are moving right to left and the positive ones are moving left to right (or vice versa, but most people whose native language is written left-to-right find it more natural to have the positive direction pointing to the right).

    So -c just means moving to the left at the speed of light and +c means moving to the right at the speed of light. Be careful to get the signs of ##u## and ##v## right as well, and it will all come out.
     
  15. Dec 4, 2015 #14
    I have no problem on understanding negative velocities (excuse me, because I didn't know the term "velocity" in english means the speed vector, as in spanish "velocidad" means speed and also its vector, althought we also call the module of the vector "celeridad").
    But getting back to the issue, given the fact that both primary equations are right, where is the sign problem then? When you say to be careful with u and v signs, where exactly do you mean? Because I can't think of a factor in the equation whose sign needs to be changed.
    If the photon is moving towards one direction (I always thought to the positive wing of OX as you well said), how can a observer watch it moving towards the other direction?

    This makes sense if the moving observer (which is the one where we have this little issue) is moving towards the negative wing of OX, so it definitely be watching the photon at -c. So, as I asked; where is the sign issue?

    Thank you for resolving my doubts and for your patience


    _______________________________________________--
    _______________________________________________--

    Final edit:

    I saw where did I get the issue. I was putting the lightspeed c in the variable v, which was, as I stated up, the speed at which observer 2 is moving. That is obviously NOT c!

    If we have, for the static observer
    [tex]u=\frac{u'+ v}{1+ \frac{u'v}{c^2}}[/tex]
    If we make u'=c
    [tex]u=\frac{c+ v}{1+ \frac{cv}{c^2}}= \frac{c+ v}{1+ \frac{v}{c}}= \frac{c(c+ v)}{c+ v}= c[/tex]

    For the moving observer, we need to put c on the u variable, not the v one. This u=c, as stated right up of this sentence. Let's go for it;

    We have, for the moving observer
    [tex]u'=\frac{u- v}{1- \frac{uv}{c^2}}[/tex]
    If u=c
    [tex]u'=\frac{c- v}{1- \frac{cv}{c^2}}= \frac{c- v}{1- \frac{v}{c}}= \frac{c(c- v)}{c- v}= c[/tex]

    Was that correct? I am imposing the condition to the first equation, but don't know if it implies some sort of math issue and I am simply forcing the result
     
    Last edited: Dec 4, 2015
  16. Dec 12, 2015 #15
    For a massive object (as opposed to massless) this equation can be written as ##E=\gamma mc^2## where ##\gamma## has the usual definition, $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$
    The liberty of calling ##\gamma m## the mass is something that Hawking is doing here, and has been done by many others in many circumstances. This includes introducing relativity to college students enrolled in introductory physics classes. But a review of the textbooks used in introductory courses reveals that the practice has dramatically fallen out of favor in the past 20 years or so. Part of the reason for that is getting the physics right.

    The reason why an author like Hawking would choose to do such a thing in a book written for the layman is of course speculation. He did say, though, that when writing A Brief History of Time hearing that each equation will cause a loss of half the readers he limited it to one. So perhaps that's at least part of the reason.

    The idea that it would take an infinite amount of energy to accelerate an object of mass ##m## to speed ##c## can be seen as a consequence of the fact that ##\gamma## approaches infinity as ##v## approaches ##c##. Thus ##\gamma m## approaches infinity because ##m## is constant.

    The Einstein mass-energy equivalence tells us that a contribution to the mass of a composite body is made by the energies of those constituents. This is much clearer when you write ##E_o=mc^2##, where ##E_o## is the rest energy.
     
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