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The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around huge Planck scale masses.

I've received the following reply in Physics Forums:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?

D=26 critical dimension

According to David Tong the first excited states of bosonic string theory have a mass given by

$$M^2=\frac{4}{\alpha'}(1-\frac{D-2}{24})$$

Thus if the dimension of spacetime $D=26$ then the masses of the bosonic strings are zero as required.

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (2.28) etc)

That's great but i would have thought we still need to somehow cancel off the intrinsic Planck scale mass of the strings due to their Planck scale length.

The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around huge Planck scale masses.

I've received the following reply in Physics Forums:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?

D=26 critical dimension

According to David Tong the first excited states of bosonic string theory have a mass given by

$$M^2=\frac{4}{\alpha'}\Big(1-\frac{D-2}{24}\Big)$$

Thus if the dimension of spacetime $D=26$ then the masses of the bosonic strings are zero as required.

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (2.28) etc)

That's great but I would have thought that we still need to somehow cancel off the intrinsic Planck scale mass of the strings due to their Planck scale length.

The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around huge Planck scale masses.

I've received the following reply in Physics Forums:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?

The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around huge Planck scale masses.

I've received the following reply in Physics Forums:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?

D=26 critical dimension

According to David Tong the first excited states of bosonic string theory have a mass given by

$$M^2=\frac{4}{\alpha'}(1-\frac{D-2}{24})$$

Thus if the dimension of spacetime $D=26$ then the masses of the bosonic strings are zero as required.

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (2.28) etc)

That's great but i would have thought we still need to somehow cancel off the intrinsic Planck scale mass of the strings due to their Planck scale length.

The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around a huge Planck scale mass.

I've received the following reply:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?

The Energy $E$ of a fundamental string due to its length $L$ goes like

$$E\sim TL$$

where string tension $T$ is given by

$$T \sim \frac{1}{l_P^2}$$

(Using natural units $\hbar=c=1$ with planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like

$$E_G \sim -\frac{G E^2}{L}$$

Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitation binding energy $E_G$ also goes like

$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

Therefore is it possible that the gravitational binding energy of a string cancels its basic energy due to its length so that the net string mass is zero?

This would allow strings to model standard model particles, whose small masses come from interaction with the Higgs field, without carrying around huge Planck scale masses.

I've received the following reply in Physics Forums:

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

My response is:

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

Is my response valid?