Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Has relativistic mass gone out of fashion?

  1. Sep 11, 2014 #1
    For a long time I have studied and read about whether the photon has mass.
    On this forum, I could find a link to a document by Gary Oas: On the abuse and use of
    the relativistic mass.
    http://arxiv.org/pdf/physics/0504110v2.pdf

    Oas' paper tells about an investigation done to 164 students:
    "Einstein also showed that the mass of an object moving at close to the speed of light, as seen by
    an outside observer, increases. [...] It explains, among other things, why the speed of light serves
    as the ultimate speed limit in the universe. Suppose you’re in a spaceship, approaching the speed
    of light. You think ‘I’ll just step on the accelerator a little harder and I’ll pass that pesky speed
    limit, no problem.’ But it won’t work: to make your craft move faster, you have to use energy;
    the more massive the spaceship, the more energy you need. And, thanks to Einstein’s special rel-
    ativity, the mass keeps increasing, so you need more and more energy to further boost the speed.
    And you’ll never quite make it. If you reached the speed of light, your spaceship, as seen by an outside observer, would have an infinite mass and it would have taken you an infinite amount of
    energy to get there."

    I remember that exactly this way I was also introduced into relativity while in school a very long time ago. Oas' paper says that in this way misconceptions develop among first time learners of relativity.


    Now I began to really wonder, can the concept of relativistic mass really be wrong or out of fashion. I am not an expert on relativity, but how is the ultimate speed limit, the speed of
    light, now explained if not with velocity dependent mass?
    If the velocity dependent mass does not prevent a massive object of achieving the speed of light,
    what is going to do it?
    The speed of a massive object cannot exceed the speed of light.
    If the relativistic mass is abandoned and we use instead the energy, the kinetic energy, and if
    there is no upper limit to how large the kinetic energy can be, doesn't this also mean that the speed of
    a massive object can exceed the speed of light?
     
    Last edited: Sep 11, 2014
  2. jcsd
  3. Sep 11, 2014 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    No, it means you would have to input infinite energy to get an object with rest mass to go at c and this remains impossible.

    There are members on this forum who would prefer to hunt you down and hurt you if you use the term "relativistic mass" but they are much to polite to do so :smile:
     
  4. Sep 11, 2014 #3
    The speed limit is imposed by spacetime itself, via the Minkowski metric. Mass is not the issue here; there is simply no mathematical way to accelerate to c.
     
  5. Sep 11, 2014 #4

    ShayanJ

    User Avatar
    Gold Member

    Hi, I'm one of those that phinds mentions. But its OK, you're safe!!!:biggrin:
    I don't know whether it is out of fashion or not, but if its not, I hope it becomes.
    In some introductory SR books the author talks about relativistic mass and its increase with speed but it never uses it. I don't know why they define it when they don't need it!!!
    Anyway, I have two arguments for you which makes relativistic mass unnecessary even here.

    1- Imagine two persons at rest w.r.t. each other. One of them shines a laser at the other one. We can be sure that the guy receives it. But if another one passes by them with a speed greater than the speed of light, according to him, the laser pulse never reaches the other guy. But the fact that the laser pulse reaches the other guy shouldn't depend on the frame of reference! So its impossible for an object to move faster than light.

    2- Imagine it is possible to send signals faster than the speed of light. That means I can send signals back in time. Now I have a device that is scheduled to send signals back in time only if it doesn't receive a signal from itself from future. But if it receives a signal, it means sent it in the future which means it didn't receive a signal. That's a paradox. Other paradoxes of this kind can be found too. So its impossible for energy to move faster than light.
     
  6. Sep 11, 2014 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not sure you can accelerate anything physically with mathematics! In any case, it's not mathematics that limits the relative velocity, it's physics.
     
  7. Sep 11, 2014 #6
    You can accelerate mathematically by successive boosting of your velocity. You can do this as many times as you like and you will never get to c. That's maths.
     
  8. Sep 11, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    But the reason the maths works that way is physics!
     
  9. Sep 11, 2014 #8

    ChrisVer

    User Avatar
    Gold Member

    Relativistic mass is not wrong.... however it's not a useful quantity for physicists, who mostly prefer talking about frame invariant quantities... when you use relativistic mass you have to specify the ref frame from which you are using it [that is in the [itex]\gamma[/itex] factor that multiplies the rest mass, and [itex]\gamma[/itex] depends on the relative velocity].
    In my opinion, I don't see how Minkowksi metric rules out velocities v>c...they exist in it but they are not causally connected. The minkowski metric gives the whole space of [itex](vec{x},t)[/itex] and from that you get timelike,spacelike and null vectors...
    It's the energy-momentum relation which does that, if for example you try to find the speed of a particle which has [itex]E^{2}=p^{2}+m^{2}, ~(c=1)[/itex] then one can find that the velocity is given by:
    [itex] |v|^2 = \frac{p^2}{E^2} [/itex] (frame independent)
    or [itex]|v|^2= \frac{p^{2}}{p^{2}+m^{2}} \le 1[/itex]
    In any case is the energy-momentum relation a result of the minkowski spacetime?
    Maybe one could try to figure that out in GR frame where he'd have:
    [itex] g_{\mu \nu} p^{\mu} p^{\nu}= m^{2} [/itex] and look at [itex]m \rightarrow 0[/itex]. However if I recall well, the results don't differ much? Plus I believe the velocities won't be the same as in SR (they'll be local)...
     
    Last edited: Sep 11, 2014
  10. Sep 11, 2014 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##c - 1 + 2 = c + 1##

    There it is. Nothing stopped me doing that! Not mathematically, anyway. What mathematics are you talking about?
     
  11. Sep 11, 2014 #10

    ShayanJ

    User Avatar
    Gold Member

    How did you get those equations?

    Anyway, I gave arguments that forbid faster than light motion and communication without using relativistic mass.

    But what I mean, isn't that relativistic mass is wrong. It has no meaning to say that its right or wrong. Its a physicist's choice to let it in the theory or not. I only mean, why should we let something in that is of no use and only causes troubles?!
     
    Last edited: Sep 11, 2014
  12. Sep 11, 2014 #11

    ChrisVer

    User Avatar
    Gold Member

    How exactly did you reach that?
    [itex] v^2 = \frac{p^2}{E^2} = \frac{p^2}{p^2 + m^2} [/itex]
    if you go the other way:
    [itex] v^2 = \frac{p^2}{E^2} = \frac{E^2-m^2}{E^2}= 1 - \frac{m^2}{E^{2}} \le 1 [/itex] again
    except for if you are asking me why [itex]v^{2} = \frac{p^2}{E^{2}}[/itex]?
    That's SR basics...
    [itex] \frac{p^{i}}{E} = \frac{\gamma m v^{i}}{\gamma m} = v^{i}[/itex]
    correction which doesn't depend on any relative velocity

    Because I didn't want to have the vector indices, I just took the inner product...
     
    Last edited: Sep 11, 2014
  13. Sep 11, 2014 #12

    ShayanJ

    User Avatar
    Gold Member

    Yeah, I got it.
    But [itex] v^i [/itex] isn't frame independent.
    The formulas [itex] \vec p=\gamma m \vec v [/itex] and [itex] E=\gamma m [/itex] contain a [itex] \gamma [/itex] and that is a function of a velocity, But what velocity? That's frame-dependent and so those formulas and fractions are frame-dependent.

    EDIT:
    Yeah, that's better!
     
  14. Sep 11, 2014 #13

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    It's very important to keep in mind that the object
    [tex]\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}[/tex]
    is not the spatial part of a three vector. It has no simple transformation properties in Minkowski space (it's of course a three vector under rotations).

    That's why one introduces the idea of proper time, which is the time in the restframe of the particle (supposed it's not massless, since massless particles have no restframe). Its increment is given as
    [tex]\mathrm{d} \tau= \mathrm{d} t \sqrt{1-\vec{v}^2/c^2}=\sqrt{c^2 \mathrm{d} t^2-\mathrm{d} \vec{x}^2}/c,[/tex]
    which shows that it is a scalar. You can write it, using an arbitrary parameter of the worldline of the particle
    [tex]\mathrm{d} \tau = \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\nu}},[/tex]
    where the dot indicates the derivative wrt. [itex]\lambda[/itex].

    Then the quantity
    [tex]u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},[/tex]
    the four-velocity of the particle, is a four-vector.

    The four-momentum is
    [tex]p^{\mu}=m u^{\mu},[/tex]
    where [itex]m[/itex] is the invariant mass of the particle, which is a scalar. The 0-component is [itex]E/c[/itex], where [itex]E[/itex] is the kinetic energy of the particle plus the rest energy [itex]E_{0}=m c^2[/itex]:
    [tex]E=p^0 c=m c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau}=m c^2 \gamma,[/tex]
    with
    [tex]\gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.[/tex]
    The momentum is
    [tex]\vec{p}=m \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} = m \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m \gamma \vec{v}.[/tex]
    The notion of a relativistic mass is unnecessary and confusing. It's not used in physicists' everyday life anymore (at least not in the high-energy particle and nuclear physics community).

    In addition it's irresponsible to teach the idea of photons in high school. It can only be properly understood in terms of relativistic quantum field theory and gauge symmetry. This is way beyond reach of high school students! The old idea of photons, introduced by Einstein in 1905, is the most misleading way to look at electromagnetic radiation. It's much better to teach it in terms of classical field theory. Most properties of em. radiation, including quantum phenomena as the photo effect can be explained using classical em. waves and quantized massive particles, and non-relativistic quantum theory is well in the reach of high-school students. The damage done with introducing these wrong concepts of the photon are obvious even at university level, when students are really able to learn the true thing. You always have to tell them to forget what they have learnt about photons before it's treated in the proper way, i.e., using QED!
     
    Last edited: Sep 11, 2014
  15. Sep 11, 2014 #14

    Nugatory

    User Avatar

    Staff: Mentor

    Speeds don't add that way (google for "relativistic velocity addition" or look for many threads here).

    If you are moving at speed ##u<c## relative to me, and you boost your speed by ##v<c##, your speed relative to me will not be ##u+v##, it will be ##(u+v)/(1+uv)## (measuring time in seconds and distances in seconds so that ##c=1## - otherwise I'd need a ##c^2## in there too).

    Try it with that math and you won't get above ##c##.
     
  16. Sep 11, 2014 #15

    ChrisVer

    User Avatar
    Gold Member

    Well that' almost right about the gamma, however the velocity in gamma refers not to the particle's velocity, but to the velocity between 2 frames which measure the same thing... (like the example with the 2 spaceships with the one launching a rocket).. the [itex]v^{i}[/itex] corresponds to the rocket's velocity as measured by one of them, and the velocity in [itex]\gamma[/itex] refers to the velocity between those two... that's why I initially called it frame independent- two frames will measure the velocity in the same way without having to correlate their measurements. However I think it sounded weird and I am somewhere wrong.
    By getting rid of it, you can have the particle's velocity by knowing its energy and momentum at your ref frame, and without having to care about the different frames.
    That's also why the quantity [itex]\frac{p^2}{E^2}
    [/itex] appears in phase integrals which are Lorentz Invariant...
     
    Last edited: Sep 11, 2014
  17. Sep 11, 2014 #16

    Ok, you did not need relativistic mass to give arguments that forbid faster than light motion and communication. You told that it is impossible for energy to move faster than light.

    Now I am thinking in layman's terms: how does energy have speed? I think this means the speed of light, and light has energy, and the speed of this energy cannot exceed the speed limit.
    I am looking for an explanation whether or not a massive object, not just energy or a photon with a zero rest mass, can travel with a speed exceeding the speed of light if the concept of relativistic mass is abandoned.

    I think you don't abandon the concept of mass altogether. I mean that there is still
    the concept called the velocity independent mass, or invariant mass, or "proper mass" or rest mass m[itex]_{0}[/itex] which can be nonzero for particles other than photons.
    If the concept of relativistic mass is abandoned, can an object or a particle with rest mass m[itex]_{0}[/itex]≠0 if it has unlimited amount of kinetic energy travel with a speed v exceeding c, in other words v>c?
     
  18. Sep 11, 2014 #17

    jtbell

    User Avatar

    Staff: Mentor

    It’s by now a well-established experimental fact that as an object moves faster, its “resistance” to going still faster, increases. Coming from a Newtonian background with F=ma, etc., it seems intuitively appealing to explain this by saying that m increases with velocity. The problem is that there’s no single way to do this mathematically that works in a wide variety of Newtonian formulas. In effect, there’s not one “relativistic mass”, but several.

    Even in the early days of relativity (early 1900s), physicists recognized this. In the early literature about relativity, you find references to “transverse mass” for F=ma when the force is perpendicular to the velocity, and “longitudinal mass” for F=ma when the force is parallel to the velocity. If the force is at some other angle, you get something in between, and the acceleration isn’t parallel to the force any more!

    It turns out that “transverse mass” also works for momentum, p=mv, but neither “mass” works for kinetic energy, K=(1/2)mv^2.

    I suspect that “transverse mass” became “the relativistic mass” because the earliest experiments dealing with fast-moving particles were mass-spectrometer type experiments, in which particles travel in a circular path in a magnetic field, under the influence of a centripetal (perpendicular) magnetic force. Also, one of the first practical effects of “relativistic mass” was in developing high-energy circular particle accelerators (synchrotrons), which similarly use perpendicular magnetic forces, and in which relativistic effects have to be taken into account, unlike the earlier low-energy cyclotrons.

    Same with me, in my second-year undergraduate “intro modern physics” course in the early 1970s. I don't remember if my high-school physics class did anything with relativity. However, when I was in graduate school in high-energy particle physics in the late 1970s and early 1980s, the only place I saw “relativistic mass” was in a textbook about accelerator design that had been written in the 1950s by someone who had worked with E. O. Lawrence on the first cyclotrons in the 1920s or 1930s, and later helped develop synchrotrons. Otherwise, when the people I worked with talked about “mass”, they always meant “rest mass” a.k.a. “invariant mass”.

    Einstein himself said in his later years that he thought it was not a good idea to use a velocity-dependent mass:

    (http://en.wikipedia.org/wiki/Mass_in_special_relativity)

    I think introductory textbooks were slow to drop “relativistic mass” because they’re usually not written by specialists, and so they tend to be conservative with changes like this. But many of them have changed. I used to teach an intro modern physics course using Beiser’s textbook. I started with (I think) the 4th edition in the late 1980s, which had a section about “relativistic mass”, and some homework problems. In the 6th edition (late 1990s or early 2000s), “relativistic mass” appeared only in a short sidebar which acknowledged that many popular-level books and some introductory textbooks still use it, but indicated that most physicists don’t.
     
    Last edited: Sep 11, 2014
  19. Sep 11, 2014 #18

    ChrisVer

    User Avatar
    Gold Member

    yes, it can theoretically (mathematically) if you allow for [itex]m^{2}_0 <0 [/itex]
    But even with infinite energy, a massive particle with [itex]m^{2}_0>0[/itex] won't reach c... it will always approach c....
     
  20. Sep 11, 2014 #19

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In a single reference frame, speeds do add like that, don't they? If I'm moving at u (in your reference frame) and I increase my speed by v (in your reference frame) then I'm moving at u + v (in your reference frame).

    Also, suppose (in my reference frame) one object is moving at c/2 in one direction and another object is moving at c/2 in the opposite direction. Then (in my reference frame) the relative velocity between the two objects is c/2 + c/2 = c.

    In any case, my point was that there's nothing mathematically stopping me adding velocities. That may not be the way the physics works in certain contexts (such as implicitly adding velocities from different reference frames!).
     
  21. Sep 11, 2014 #20

    ChrisVer

    User Avatar
    Gold Member

    this is not how the relativistic velocity addition is done though... by your idea, if you have something moving with +c and something moving with -c, then their relative velocity will be c+c=2c?!!?
    You have to understand that the way you add velocities in the Gallilean way (by simply adding them up) is just a low-velocity approximation of how it really works.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Has relativistic mass gone out of fashion?
  1. Relativistic Mass (Replies: 16)

  2. Relativistic Mass (Replies: 12)

  3. Relativistic mass? (Replies: 17)

  4. Relativistic mass (Replies: 32)

  5. Relativistic mass (Replies: 28)

Loading...