E Vector in E=mc2? Scalar Multiplication & Vectors

In summary, energy is a scalar quantity and not a vector. The formula for potential energy, U=mgh, involves vectors such as the gravitational force, but the end result is a scalar. This can also be seen in the formula for kinetic energy, which is a scalar. In relativity, energy is a component of the energy-momentum four vector, but it is still a scalar. The relationship between a force and its associated potential energy also shows that energy must be a scalar. The units of energy, such as the joule, do not determine whether it is a vector or scalar. The h in the formula for potential energy is a scalar, representing distance. Energy can be transferred between objects, but this does not make it a
  • #1
andrecoelho
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TL;DR Summary
E=mc2 , E vector?
If the energy is a vector, which as i understand for example, the potential energy , U=mgh, where g is the gravitational force, Then U is the product of scalars and vectors, so its a vector
In that case being E a vector , can it be equal to mc2 (each are scalars). Like mulitplication of scalars are not vectors...
 
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  • #3
But isn't energy the exorthation of force in a body?
 
  • #4
I have no idea what “exorthation” means. But regardless of what that word means, energy is not a vector. Force is a vector, but that doesn’t mean that everything that has something vaguely to do with force is also a vector.
 
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  • #5
Dale said:
I have no idea what “exorthation” means. But regardless of what that word means, energy is not a vector.
exorthaton -> the effect of applying the force

Can you explain why? Particulary the U=mgh formula
 
  • #7
andrecoelho said:
Summary:: E=mc2 , E vector?

If the energy is a vector, which as i understand for example, the potential energy , U=mgh, where g is the gravitational force, Then U is the product of scalars and vectors, so its a vector
In that case being E a vector , can it be equal to mc2 (each are scalars). Like mulitplication of scalars are not vectors...

In [itex] U=mgh [/itex], the [itex] g [/itex] is not the gravitational force--- [itex] m\vec g [/itex] is the gravitational force (and [itex] \vec g [/itex] is the gravitational field).
More completely, [itex] U=-(m\vec g)\cdot \vec h [/itex] (as minus the work done by the gravitational force).
The dot-product of two vectors is a scalar. So, [itex] U [/itex] is a scalar.
In fact, the gravitational potential [itex] gy [/itex] is a scalar field
(so minus its gradient is the gravitational field)
and so the gravitational potential energy [itex] mgy [/itex] is a scalar.

Note that ordinary kinetic energy (from the work-energy theorem) is a scalar [itex] K=\frac{1}{2}mv^2 [/itex].

Energy is a scalar.
 
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  • #8
andrecoelho said:
exorthaton -> the effect of applying the force

Can you explain why? Particulary the U=mgh formula
The ##g## in that formula is the magnitude of the gravitational force: ##g = |\vec g|##. More generally, we have $$U = -\frac{GMm}{r}$$
In any case, potential energy is not a vector; nor is any form of energy. Kinetic energy is ##\frac 1 2 mv^2##, where again ##v = |\vec v|## is the magnitude of the velocity. So, KE is also a not a vector.

As you've posted this in relativity forum, the energy of a particle is a component of the energy-momentum four vector: $$\mathbf{p} = (E, p_x, p_y, p_z) = (E, \vec p)$$ In that sense, energy is a component of a four-vector.
 
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Keeping ##c## it's
$$(p^{\mu})=(E/c,\vec{p}).$$
One should also use invariant masses and only invariant masses in relativity, and thus the relation between energy and momentum of a particle reads
$$p_{\mu} p^{\mu}=\frac{E^2}{c^2}+\vec{p}^2=m^2 c^2$$
or
$$E=c \sqrt{(m c)^2+\vec{p}^2}.$$
In terms of the coordinate velocity ##\vec{v}=\vec{p} c^2/E## it reads
$$E=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}.$$
The famous formula by Einstein should read (BTW also according to Einstein himself!)
$$E_0=m c^2,$$
where ##E_0## is the energy as measured in the rest frame of the particle.
 
  • #10
andrecoelho said:
Can you explain why? Particulary the U=mgh formula
In general a force and its associated potential energy have the relationship: ##\vec F = -\nabla U##. Since ##\nabla## is an operator that converts a scalar into a vector it is clear that ##U## must always be a scalar.

In the case of a uniform gravitational field we have $$\vec F = m \vec g = -\nabla (mgh) = -\nabla U$$ where on the left hand side ##\vec g## is a vector pointing downwards and on the right hand side ##g## is the magnitude of that vector.
 
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  • #11
What is the relevance of attributing a unit to a expression? does that mandate that the quantity is vectorial?!
what is the relation between a joule and the fact that its possible a vector?
And i still don't get it, h is not a vector , is distance
" If you accept that energy, by definition, is a scalar function," , i can't just assume that something just by defintion, its not an axiom. If the math is incorrect, its also incorrect in terms of physics, because physics use exclusive maath
is the h in the formula a vector (lisnt length a scalar)??
Energy can be transferred one side to another, hence could be a vector the problem remains,
is energy released to every part of a body, and each part is the same? or is there a parts of the body with more energy
than others? Just because we don't know what happens to energy after being transferred (for example The force used to store energy in a rubber band has a certain direction,
that doesn't mean its not a vector, so one thing is sure, energy can be trasferred (which applies being vector) . If you don't know where the energy goes,
how can you know its not a vector

Does the energy gets released equally?
or the whole body has energy (where other parts dont). Does energy already remeains at a body and energy gets added
beign released means going into a direction or several
 
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  • #13
andrecoelho said:
What is the relevance of attributing a unit to a expression? does that mandate that the quantity is vectorial?!
No, assigning units in no way makes the quantity a vector. I am not sure what would possibly lead you to believe that.

andrecoelho said:
i can't just assume that something just by defintion
Why not? Definitions are true by definition.

andrecoelho said:
Energy can be transferred one side to another, hence could be a vector the problem remains,
Actually, no. What you are thinking of is energy flux, not energy. Energy flux density is a vector.

andrecoelho said:
If you don't know where the energy goes,
how can you know its not a vector
We know that it is not a vector because it is defined to not be a vector. Since it is a definition it is true by definition.

andrecoelho said:
energy can be trasferred (which applies being vector)
This is nonsense. Money can also be transferred. Money is not a vector. Transferred applies more to conserved quantities than to vector quantities.

There really is no way around this and absolutely no ambiguity on this topic whatsoever. Energy is not a vector.

Since there is literally nothing more to say on this topic and since you seem disinclined to actually learn anything here, this thread is closed.
 
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FAQ: E Vector in E=mc2? Scalar Multiplication & Vectors

What is the significance of the "E" vector in the equation E=mc2?

The "E" vector in the equation E=mc2 represents the energy of an object. It is a vector quantity because it has both magnitude and direction, and it is directly proportional to the mass of the object.

How is scalar multiplication related to vectors in E=mc2?

In E=mc2, scalar multiplication refers to the "c2" term, which is the speed of light squared. This scalar value is multiplied by the mass of the object to determine the amount of energy it contains.

What is the difference between scalar and vector quantities in E=mc2?

In E=mc2, scalar quantities are represented by the "c2" term and have only magnitude, while vector quantities are represented by the "E" vector and have both magnitude and direction.

How does the "E" vector in E=mc2 relate to Einstein's theory of relativity?

The "E" vector in E=mc2 is a fundamental component of Einstein's theory of relativity, which states that energy and mass are equivalent and can be converted into each other. This equation shows the relationship between the two and the role of the speed of light in this conversion.

Can the "E" vector in E=mc2 be negative?

Yes, the "E" vector in E=mc2 can be negative. This indicates that the object has negative energy, which is possible in certain scenarios such as antimatter reactions. However, in most cases, the "E" vector will be positive, representing the object's energy in a given direction.

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