The equation is: $\lambda = \frac{h}{mv}$. m here means the mass of the object while moving or while stationary?

  • $\begingroup$ It is better to write the deBroglie equation as $\lambda=\frac{h}{p}$, where p is the momentum. If the motion is relativistic then $p=\gamma mv$ where m is the rest mass. $\endgroup$ Commented Oct 17, 2017 at 5:26
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    $\begingroup$ Possible duplicate of De Broglie wavelength, frequency and velocity - interpretation $\endgroup$
    – Masa
    Commented Oct 17, 2017 at 5:32
  • $\begingroup$ Why do people here are so sensitive about "duplicate"? my question and the question Masoud suggested as duplicate may have something in common but they are not the same. what is the problem in just answering the question instead of searching the lot to find a question that MAY BE duplicate!! Two questions are asked from two different point of view. May their answer have something in common, but they are not the same question!! $\endgroup$
    – Touhid
    Commented Oct 17, 2017 at 5:37
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    $\begingroup$ @Touhid it's usefull in principle to link similar questions in the comments. If your question were an exact duplicate or a duplicate to an extent that this forum discourages, the question would be closed, where having similar questions linked facilitates this process. $\endgroup$ Commented Oct 17, 2017 at 6:11
  • $\begingroup$ That said, the question linked as a "possible duplicate" is not actually a duplicate of this particular question, as it doesn't deal with relativistic mass at all. $\endgroup$ Commented Oct 17, 2017 at 14:00

1 Answer 1


The de Broglie wavelength is given by:

$$ \lambda = \frac{h}{p} $$

where $p$ is the momentum. Under non-relativistic conditions the momentum is just:

$$ p = mv $$

which gives the equation you cited. At relativistic speeds the momentum is given by:

$$ p = \gamma mv $$

where $\gamma$ is the Lorentz factor given by:

$$ \gamma = \sqrt{1 - \frac{v^2}{c^2}} $$

In these equations $m$ is the rest mass and this is an invariant and not dependent on the speed.

  • $\begingroup$ how do we get p=γmv? $\endgroup$
    – Touhid
    Commented Oct 17, 2017 at 14:06
  • $\begingroup$ @Touhid: when you formulate mechanics in a relativistic way the momentum is replaced by a four-vector called the four momentum. The $\gamma mv$ is the spatial part of this four vector. $\endgroup$ Commented Oct 17, 2017 at 14:15

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